Solutions to Problems

3.1 Solutions to Problems in Special Relativity Theory

1.1
a) A scalar is always the same independent of the inertial frame. The scalar product between the two 4 -vectors A → A → vec(A)\vec{A}A→ and B → B → vec(B)\vec{B}B→ is zero in the inertial frame of the observer O ′ O ′ O^(')\mathcal{O}^{\prime}O′ since they point in different coordinate directions. Thus, it is always zero. The two 4 -vectors U → U → vec(U)\vec{U}U→ and V → V → vec(V)\vec{V}V→ are lightlike in different directions. The scalar product between two lightlike 4 -vectors pointing in different directions is always nonzero. Assume that U 0 = | α | , U 1 = − | α | U 0 = | α | , U 1 = − | α | U^(0)=|alpha|,U^(1)=-|alpha|U^{0}=|\alpha|, U^{1}=-|\alpha|U0=|α|,U1=−|α| and V 0 = | β | , V 1 = | β | V 0 = | β | , V 1 = | β | V^(0)=|beta|,V^(1)=|beta|V^{0}=|\beta|, V^{1}=|\beta|V0=|β|,V1=|β|, where α , β ≠ 0 α , β ≠ 0 alpha,beta!=0\alpha, \beta \neq 0α,β≠0, then we have
(3.1) U → ⋅ V → = η μ ν U μ V ν = η 00 U 0 V 0 + η 11 U 1 V 1 = | α | | β | − ( − | α | ) | β | = 2 | α β | > 0 . (3.1) U → ⋅ V → = η μ ν U μ V ν = η 00 U 0 V 0 + η 11 U 1 V 1 = | α | | β | − ( − | α | ) | β | = 2 | α β | > 0 . {:(3.1) vec(U)* vec(V)=eta_(mu nu)U^(mu)V^(nu)=eta_(00)U^(0)V^(0)+eta_(11)U^(1)V^(1)=|alpha||beta|-(-|alpha|)|beta|=2|alpha beta| > 0.:}\begin{equation*} \vec{U} \cdot \vec{V}=\eta_{\mu \nu} U^{\mu} V^{\nu}=\eta_{00} U^{0} V^{0}+\eta_{11} U^{1} V^{1}=|\alpha||\beta|-(-|\alpha|)|\beta|=2|\alpha \beta|>0 . \tag{3.1} \end{equation*}(3.1)U→⋅V→=ημνUμVν=η00U0V0+η11U1V1=|α||β|−(−|α|)|β|=2|αβ|>0.
In conclusion, the statements 1,2 , and 6 are true.
b) The 4 -vector A → A → vec(A)\vec{A}A→ can be proportional to a 4 -velocity vector for a massive particle, since A → A → vec(A)\vec{A}A→ is equal to ( A 0 , 0 , 0 , 0 ) A 0 , 0 , 0 , 0 (A^(0),0,0,0)\left(A^{0}, 0,0,0\right)(A0,0,0,0) in S ′ S ′ S^(')S^{\prime}S′ a 4-velocity vector of a massive particle is given by ( c , 0 , 0 , 0 ) ( c , 0 , 0 , 0 ) (c,0,0,0)(c, 0,0,0)(c,0,0,0), where c c ccc is the speed of light. The two 4 -vectors U → U → vec(U)\vec{U}U→ and V → V → vec(V)\vec{V}V→ are proportional to the worldline tangent vector of massless particles, since they are null vectors. Vector B → B → vec(B)\vec{B}B→ is spacelike and cannot be the tangent of a worldline.

1.2

a) Let A μ A μ A^(mu)A^{\mu}Aμ be the timelike 4 -vector. There exists an inertial coordinate system K ′ K ′ K^(')K^{\prime}K′ such that A ′ μ = ( A ′ 0 , 0 ) A ′ μ = A ′ 0 , 0 A^('mu)=(A^('0),0)A^{\prime \mu}=\left(A^{\prime 0}, 0\right)A′μ=(A′0,0), where A ′ 0 ≠ 0 A ′ 0 ≠ 0 A^('0)!=0A^{\prime 0} \neq 0A′0≠0. If the 4 -vector B μ B μ B^(mu)B^{\mu}Bμ is orthogonal to A μ A μ A^(mu)A^{\mu}Aμ, then A ′ μ B μ ′ = A ′ 0 B 0 ′ = 0 A ′ μ B μ ′ = A ′ 0 B 0 ′ = 0 A^('mu)B_(mu)^(')=A^('0)B_(0)^(')=0A^{\prime \mu} B_{\mu}^{\prime}=A^{\prime 0} B_{0}^{\prime}=0A′μBμ′=A′0B0′=0, which means that B 0 ′ = B 0 = 0 B 0 ′ = B 0 = 0 B_(0)^(')=B^(0)=0B_{0}^{\prime}=B^{0}=0B0′=B0=0. Hence, it holds that
(3.2) B 2 = B ′ 2 = B ′ μ B μ ′ = − B ′ 2 < 0 (3.2) B 2 = B ′ 2 = B ′ μ B μ ′ = − B ′ 2 < 0 {:(3.2)B^(2)=B^('2)=B^('mu)B_(mu)^(')=-B^('2) < 0:}\begin{equation*} B^{2}=B^{\prime 2}=B^{\prime \mu} B_{\mu}^{\prime}=-B^{\prime 2}<0 \tag{3.2} \end{equation*}(3.2)B2=B′2=B′μBμ′=−B′2<0
which means that B μ B μ B^(mu)B^{\mu}Bμ is spacelike. (We assume that B μ ≢ 0 B μ ≢ 0 B^(mu)≢0B^{\mu} \not \equiv 0Bμ≢0.)
b) Let A μ A μ A^(mu)A^{\mu}Aμ and B μ B μ B^(mu)B^{\mu}Bμ be the 4 -vectors. There exists an inertial coordinate system K ′ K ′ K^(')K^{\prime}K′ such that A ′ μ = ( A ′ 0 , 0 ) A ′ μ = A ′ 0 , 0 A^('mu)=(A^('0),0)A^{\prime \mu}=\left(A^{\prime 0}, \mathbf{0}\right)A′μ=(A′0,0) and B ′ μ = ( B ′ 0 , B ′ ) B ′ μ = B ′ 0 , B ′ B^('mu)=(B^('0),B^('))B^{\prime \mu}=\left(B^{\prime 0}, B^{\prime}\right)B′μ=(B′0,B′), where A ′ 0 > 0 A ′ 0 > 0 A^('0) > 0A^{\prime 0}>0A′0>0 and B 0 > 0 B 0 > 0 B^(0) > 0B^{0}>0B0>0. Therefore, it holds that
( A ′ μ + B ′ μ ) ( A μ ′ + B μ ′ ) = ( A ′ 0 + B ′ 0 ) 2 − B ′ 2 (3.3) = ( A ′ 0 ) 2 + 2 A ′ 0 B ′ 0 + ( B ′ 0 ) 2 − B ′ 2 > 0 A ′ μ + B ′ μ A μ ′ + B μ ′ = A ′ 0 + B ′ 0 2 − B ′ 2 (3.3) = A ′ 0 2 + 2 A ′ 0 B ′ 0 + B ′ 0 2 − B ′ 2 > 0 {:[(A^('mu)+B^('mu))(A_(mu)^(')+B_(mu)^('))=(A^('0)+B^('0))^(2)-B^('2)],[(3.3)=(A^('0))^(2)+2A^('0)B^('0)+(B^('0))^(2)-B^('2) > 0]:}\begin{align*} \left(A^{\prime \mu}+B^{\prime \mu}\right)\left(A_{\mu}^{\prime}+B_{\mu}^{\prime}\right) & =\left(A^{\prime 0}+B^{\prime 0}\right)^{2}-B^{\prime 2} \\ & =\left(A^{\prime 0}\right)^{2}+2 A^{\prime 0} B^{\prime 0}+\left(B^{\prime 0}\right)^{2}-B^{\prime 2}>0 \tag{3.3} \end{align*}(A′μ+B′μ)(Aμ′+Bμ′)=(A′0+B′0)2−B′2(3.3)=(A′0)2+2A′0B′0+(B′0)2−B′2>0
because ( B ′ 0 ) 2 − B ′ 2 > 0 B ′ 0 2 − B ′ 2 > 0 (B^('0))^(2)-B^('2) > 0\left(B^{\prime 0}\right)^{2}-B^{\prime 2}>0(B′0)2−B′2>0, since B μ B μ B^(mu)B^{\mu}Bμ is timelike.
c) There exists an inertial coordinate system K ′ K ′ K^(')K^{\prime}K′ such that the spacelike 4-vector has the components A ′ μ = ( 0 , A ′ 1 , 0 , 0 ) A ′ μ = 0 , A ′ 1 , 0 , 0 A^('mu)=(0,A^('1),0,0)A^{\prime \mu}=\left(0, A^{\prime 1}, 0,0\right)A′μ=(0,A′1,0,0) relative to K ′ K ′ K^(')K^{\prime}K′. Thus, we can set
(3.4) A ′ μ = 1 2 ( A ′ 1 , A ′ 1 , 0 , 0 ) − 1 2 ( A ′ 1 , − A ′ 1 , 0 , 0 ) (3.4) A ′ μ = 1 2 A ′ 1 , A ′ 1 , 0 , 0 − 1 2 A ′ 1 , − A ′ 1 , 0 , 0 {:(3.4)A^('mu)=(1)/(2)(A^('1),A^('1),0,0)-(1)/(2)(A^('1),-A^('1),0,0):}\begin{equation*} A^{\prime \mu}=\frac{1}{2}\left(A^{\prime 1}, A^{\prime 1}, 0,0\right)-\frac{1}{2}\left(A^{\prime 1},-A^{\prime 1}, 0,0\right) \tag{3.4} \end{equation*}(3.4)A′μ=12(A′1,A′1,0,0)−12(A′1,−A′1,0,0)
d) Introduce K ′ K ′ K^(')K^{\prime}K′ in the same way as in b). Then, it holds that
(3.5) A μ B μ = A ′ μ B μ ′ = A ′ 0 B 0 ′ > 0 (3.5) A μ B μ = A ′ μ B μ ′ = A ′ 0 B 0 ′ > 0 {:(3.5)A^(mu)B_(mu)=A^('mu)B_(mu)^(')=A^('0)B_(0)^(') > 0:}\begin{equation*} A^{\mu} B_{\mu}=A^{\prime \mu} B_{\mu}^{\prime}=A^{\prime 0} B_{0}^{\prime}>0 \tag{3.5} \end{equation*}(3.5)AμBμ=A′μBμ′=A′0B0′>0
1.3
The relative gamma factor between two observers is given by
(3.6) γ 12 = V 1 ⋅ V 2 (3.6) γ 12 = V 1 ⋅ V 2 {:(3.6)gamma_(12)=V_(1)*V_(2):}\begin{equation*} \gamma_{12}=V_{1} \cdot V_{2} \tag{3.6} \end{equation*}(3.6)γ12=V1⋅V2
where V 1 V 1 V_(1)V_{1}V1 and V 2 V 2 V_(2)V_{2}V2 are the 4 -velocities of the observers. (This may be seen by going to the rest frame of one of the observers, where the 4 -velocity of two other is given by V = γ 12 ( 1 , v rel ) V = γ 12 1 , v rel  V=gamma_(12)(1,v_("rel "))V=\gamma_{12}\left(1, \boldsymbol{v}_{\text {rel }}\right)V=γ12(1,vrel ).) We find that
(3.7) V 1 = γ 1 ( 1 , v 1 ) , V 2 = γ 2 ( 1 , v 2 ) (3.7) V 1 = γ 1 1 , v 1 , V 2 = γ 2 1 , v 2 {:(3.7)V_(1)=gamma_(1)(1,v_(1))","quadV_(2)=gamma_(2)(1,v_(2)):}\begin{equation*} V_{1}=\gamma_{1}\left(1, \boldsymbol{v}_{1}\right), \quad V_{2}=\gamma_{2}\left(1, \boldsymbol{v}_{2}\right) \tag{3.7} \end{equation*}(3.7)V1=γ1(1,v1),V2=γ2(1,v2)
where
(3.8) γ i = 1 1 − v i 2 (3.8) γ i = 1 1 − v i 2 {:(3.8)gamma_(i)=(1)/(sqrt(1-v_(i)^(2))):}\begin{equation*} \gamma_{i}=\frac{1}{\sqrt{1-v_{i}^{2}}} \tag{3.8} \end{equation*}(3.8)γi=11−vi2
Consequently, we obtain
(3.9) γ 12 = V 1 ⋅ V 2 = γ 1 γ 2 ( 1 − v 1 ⋅ v 2 ) = 1 1 − v 1 2 1 − v 2 2 ( 1 − v 1 ⋅ v 2 ) (3.9) γ 12 = V 1 ⋅ V 2 = γ 1 γ 2 1 − v 1 ⋅ v 2 = 1 1 − v 1 2 1 − v 2 2 1 − v 1 ⋅ v 2 {:(3.9)gamma_(12)=V_(1)*V_(2)=gamma_(1)gamma_(2)(1-v_(1)*v_(2))=(1)/(sqrt(1-v_(1)^(2))sqrt(1-v_(2)^(2)))(1-v_(1)*v_(2)):}\begin{equation*} \gamma_{12}=V_{1} \cdot V_{2}=\gamma_{1} \gamma_{2}\left(1-\boldsymbol{v}_{1} \cdot \boldsymbol{v}_{2}\right)=\frac{1}{\sqrt{1-v_{1}^{2}} \sqrt{1-v_{2}^{2}}}\left(1-\boldsymbol{v}_{1} \cdot \boldsymbol{v}_{2}\right) \tag{3.9} \end{equation*}(3.9)γ12=V1⋅V2=γ1γ2(1−v1⋅v2)=11−v121−v22(1−v1⋅v2)
1.4
a) The correct answer is: Statement 3 (never). This is due to the fact that a photon is traveling at the speed of light relative to all inertial frames. Hence, it is impossible to perform a Lorentz transformation to a frame where it is does not move.
b) The correct answer is: Statement 2 (sometimes). In order to have a rest frame for the center of momentum for a system of two photons, it cannot be moving with
the speed of light. This happens when the two photons are moving in directions parallel to each other. However, in all other configurations, the center of momentum will move at a speed slower than c c ccc, and then, we can find a rest frame for it.

1.5

a) The formula for length contraction (or Lorentz contraction) in special relativity is given by
(3.10) ℓ ′ = ℓ γ ( v ) , γ ( v ) ≡ 1 1 − v 2 c 2 (3.10) ℓ ′ = ℓ γ ( v ) , γ ( v ) ≡ 1 1 − v 2 c 2 {:(3.10)ℓ^(')=(ℓ)/(gamma(v))","quad gamma(v)-=(1)/(sqrt(1-(v^(2))/(c^(2)))):}\begin{equation*} \ell^{\prime}=\frac{\ell}{\gamma(v)}, \quad \gamma(v) \equiv \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \tag{3.10} \end{equation*}(3.10)ℓ′=ℓγ(v),γ(v)≡11−v2c2
where ℓ ℓ ℓ\ellℓ is the length of an object in its rest frame O O O\mathcal{O}O and ℓ ′ ℓ ′ ℓ^(')\ell^{\prime}ℓ′ is the corresponding length of the object in a frame O ′ O ′ O^(')\mathcal{O}^{\prime}O′ moving with constant velocity v ( 0 ≤ v < c ) v ( 0 ≤ v < c ) v(0 <= v < c)v(0 \leq v<c)v(0≤v<c) relative to the rest frame O O O\mathcal{O}O of the object. Note that 0 < ℓ ′ ≤ ℓ 0 < ℓ ′ ≤ ℓ 0 < ℓ^(') <= ℓ0<\ell^{\prime} \leq \ell0<ℓ′≤ℓ, since 1 ≤ γ ( v ) < ∞ 1 ≤ γ ( v ) < ∞ 1 <= gamma(v) < oo1 \leq \gamma(v)<\infty1≤γ(v)<∞, so the object appears to be shortened in the frame O ′ O ′ O^(')\mathcal{O}^{\prime}O′ moving relative to the rest frame O O O\mathcal{O}O.
The length contraction formula can be derived using the fact that the length of the object in a given frame is the spatial distance between the ends of the object measured at the same time in that frame. Consider the time t ′ t ′ t^(')t^{\prime}t′ in S ′ S ′ S^(')S^{\prime}S′ to be the time at which we measure the position of the ends of the object, which are then separated by ℓ ′ ℓ ′ ℓ^(')\ell^{\prime}ℓ′. Calling the events at the ends of the object A A AAA and B B BBB, respectively, the spatial distance between them in S ′ S ′ S^(')S^{\prime}S′ is therefore ℓ ′ ℓ ′ ℓ^(')\ell^{\prime}ℓ′ and the temporal separation is zero. At the same time, the spatial and temporal separations between the events in S S SSS are ℓ ℓ ℓ\ellℓ and Δ t Δ t Delta t\Delta tΔt, respectively, as the object is at rest in S S SSS. For the events to be simultaneous in S ′ S ′ S^(')S^{\prime}S′, the Lorentz transformation requires
(3.11) c Δ t ′ = γ ( c Δ t − v ℓ c ) = 0 (3.11) c Δ t ′ = γ c Δ t − v ℓ c = 0 {:(3.11)c Deltat^(')=gamma(c Delta t-(vℓ)/(c))=0:}\begin{equation*} c \Delta t^{\prime}=\gamma\left(c \Delta t-\frac{v \ell}{c}\right)=0 \tag{3.11} \end{equation*}(3.11)cΔt′=γ(cΔt−vℓc)=0
and therefore,
(3.12) Δ t = v ℓ c 2 (3.12) Δ t = v ℓ c 2 {:(3.12)Delta t=(vℓ)/(c^(2)):}\begin{equation*} \Delta t=\frac{v \ell}{c^{2}} \tag{3.12} \end{equation*}(3.12)Δt=vℓc2
As the spacetime interval between the events is invariant, we now find that
(3.13) ℓ ′ 2 − c 2 Δ t ′ 2 = ℓ ′ 2 = ℓ 2 − c 2 Δ t 2 = ℓ 2 − v 2 c 2 ⋅ ℓ 2 (3.13) ℓ ′ 2 − c 2 Δ t ′ 2 = ℓ ′ 2 = ℓ 2 − c 2 Δ t 2 = ℓ 2 − v 2 c 2 ⋅ ℓ 2 {:(3.13)ℓ^('2)-c^(2)Deltat^('2)=ℓ^('2)=ℓ^(2)-c^(2)Deltat^(2)=ℓ^(2)-(v^(2))/(c^(2))*ℓ^(2):}\begin{equation*} \ell^{\prime 2}-c^{2} \Delta t^{\prime 2}=\ell^{\prime 2}=\ell^{2}-c^{2} \Delta t^{2}=\ell^{2}-\frac{v^{2}}{c^{2}} \cdot \ell^{2} \tag{3.13} \end{equation*}(3.13)ℓ′2−c2Δt′2=ℓ′2=ℓ2−c2Δt2=ℓ2−v2c2⋅ℓ2
or, once simplified,
(3.14) ℓ ′ = ℓ 1 − v 2 c 2 = ℓ γ (3.14) ℓ ′ = ℓ 1 − v 2 c 2 = ℓ γ {:(3.14)ℓ^(')=ℓsqrt(1-(v^(2))/(c^(2)))=(ℓ)/(gamma):}\begin{equation*} \ell^{\prime}=\ell \sqrt{1-\frac{v^{2}}{c^{2}}}=\frac{\ell}{\gamma} \tag{3.14} \end{equation*}(3.14)ℓ′=ℓ1−v2c2=ℓγ
which is the length contraction formula.
b) The formula for time dilation in special relativity is given by
(3.15) t = γ ( v ) t ′ , γ ( v ) ≡ 1 1 − v 2 c 2 (3.15) t = γ ( v ) t ′ , γ ( v ) ≡ 1 1 − v 2 c 2 {:(3.15)t=gamma(v)t^(')","quad gamma(v)-=(1)/(sqrt(1-(v^(2))/(c^(2)))):}\begin{equation*} t=\gamma(v) t^{\prime}, \quad \gamma(v) \equiv \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \tag{3.15} \end{equation*}(3.15)t=γ(v)t′,γ(v)≡11−v2c2
where t t ttt is the time in the frame O O O\mathcal{O}O and t ′ t ′ t^(')t^{\prime}t′ is the corresponding time in the frame O ′ O ′ O^(')\mathcal{O}^{\prime}O′ moving with constant velocity v ( 0 ≤ v < c ) v ( 0 ≤ v < c ) v(0 <= v < c)v(0 \leq v<c)v(0≤v<c) relative to the frame O O O\mathcal{O}O. Note that t > t ′ t > t ′ t > t^(')t>t^{\prime}t>t′, since 1 ≤ γ ( v ) < ∞ 1 ≤ γ ( v ) < ∞ 1 <= gamma(v) < oo1 \leq \gamma(v)<\infty1≤γ(v)<∞, so an observer in the frame O O O\mathcal{O}O measures a longer time than an observer in O ′ O ′ O^(')\mathcal{O}^{\prime}O′ measures between two events which occur at the same spatial point in the frame O ′ O ′ O^(')\mathcal{O}^{\prime}O′.
In order to derive the time dilation formula, we consider the worldline of an observer at rest in S ′ S ′ S^(')S^{\prime}S′. In S ′ S ′ S^(')S^{\prime}S′, the spatial and temporal differences between two events on the worldline are given by zero and t ′ t ′ t^(')t^{\prime}t′, respectively. In the frame S S SSS, the observer is instead moving at speed v v vvv and the spatial and temporal separations are therefore t t ttt and v t v t vtv tvt, respectively. As the spacetime interval between the events is invariant, we find that
(3.16) c 2 t ′ 2 − 0 2 = c 2 t 2 − v 2 t 2 ⟹ t ′ = t 1 − v 2 c 2 (3.16) c 2 t ′ 2 − 0 2 = c 2 t 2 − v 2 t 2 ⟹ t ′ = t 1 − v 2 c 2 {:(3.16)c^(2)t^('2)-0^(2)=c^(2)t^(2)-v^(2)t^(2)quad Longrightarrowquadt^(')=tsqrt(1-(v^(2))/(c^(2))):}\begin{equation*} c^{2} t^{\prime 2}-0^{2}=c^{2} t^{2}-v^{2} t^{2} \quad \Longrightarrow \quad t^{\prime}=t \sqrt{1-\frac{v^{2}}{c^{2}}} \tag{3.16} \end{equation*}(3.16)c2t′2−02=c2t2−v2t2⟹t′=t1−v2c2
or, equivalently,
(3.17) t = γ ( v ) t ′ (3.17) t = γ ( v ) t ′ {:(3.17)t=gamma(v)t^('):}\begin{equation*} t=\gamma(v) t^{\prime} \tag{3.17} \end{equation*}(3.17)t=γ(v)t′
which is the time dilation formula.

1.6

Length contraction in the x x xxx-direction yields (see Figure 3.1)
(3.18) a ′ = a 1 − v 2 c 2 = { v = 2 / 3 c } = a 1 − 2 3 = a 3 , (3.18) a ′ = a 1 − v 2 c 2 = { v = 2 / 3 c } = a 1 − 2 3 = a 3 , {:(3.18)a^(')=asqrt(1-(v^(2))/(c^(2)))={v=sqrt(2//3)c}=asqrt(1-(2)/(3))=(a)/(sqrt3)",":}\begin{equation*} a^{\prime}=a \sqrt{1-\frac{v^{2}}{c^{2}}}=\{v=\sqrt{2 / 3} c\}=a \sqrt{1-\frac{2}{3}}=\frac{a}{\sqrt{3}}, \tag{3.18} \end{equation*}(3.18)a′=a1−v2c2={v=2/3c}=a1−23=a3,
where a = 1 / 2 m a = 1 / 2 m a=1//sqrt2ma=1 / \sqrt{2} \mathrm{~m}a=1/2 m. Note that there is no length contraction in the y y yyy-direction.


v → v → vec(v)\vec{v}v→
Figure 3.1 A rod inclined 45 ∘ 45 ∘ 45^(@)45^{\circ}45∘ in the x y x y xyx yxy-plane with respect to the x x xxx-axis and the angle θ θ theta\thetaθ in the x ′ y ′ x ′ y ′ x^(')y^(')x^{\prime} y^{\prime}x′y′-plane with respect to the x ′ x ′ x^(')x^{\prime}x′-axis. The relative speed between the two planes is v v vvv.
Now, Pythagoras' theorem gives
ℓ = a ′ 2 + a 2 = { a ′ = a / 3 } = a 2 3 + a 2 = 2 a 3 = { a = 1 / 2 m } = 2 3 m ℓ = a ′ 2 + a 2 = a ′ = a / 3 = a 2 3 + a 2 = 2 a 3 = { a = 1 / 2 m } = 2 3 m ℓ=sqrt(a^('2)+a^(2))={a^(')=a//sqrt3}=sqrt((a^(2))/(3)+a^(2))=(2a)/(sqrt3)={a=1//sqrt2m}=sqrt((2)/(3))m\ell=\sqrt{a^{\prime 2}+a^{2}}=\left\{a^{\prime}=a / \sqrt{3}\right\}=\sqrt{\frac{a^{2}}{3}+a^{2}}=\frac{2 a}{\sqrt{3}}=\{a=1 / \sqrt{2} \mathrm{~m}\}=\sqrt{\frac{2}{3}} \mathrm{~m}ℓ=a′2+a2={a′=a/3}=a23+a2=2a3={a=1/2 m}=23 m.
Thus, the angle θ θ theta\thetaθ is given by
(3.20) θ = arctan ⁡ a a ′ = { a ′ = a / 3 } = arctan ⁡ 3 = π 3 = 60 ∘ (3.20) θ = arctan ⁡ a a ′ = a ′ = a / 3 = arctan ⁡ 3 = π 3 = 60 ∘ {:(3.20)theta=arctan((a)/(a^(')))={a^(')=a//sqrt3}=arctan sqrt3=(pi)/(3)=60^(@):}\begin{equation*} \theta=\arctan \frac{a}{a^{\prime}}=\left\{a^{\prime}=a / \sqrt{3}\right\}=\arctan \sqrt{3}=\frac{\pi}{3}=60^{\circ} \tag{3.20} \end{equation*}(3.20)θ=arctan⁡aa′={a′=a/3}=arctan⁡3=π3=60∘
In conclusion, the answer is ℓ = 2 3 m ℓ = 2 3 m ℓ=sqrt((2)/(3))m\ell=\sqrt{\frac{2}{3}} \mathrm{~m}ℓ=23 m and θ = 60 ∘ θ = 60 ∘ theta=60^(@)\theta=60^{\circ}θ=60∘.
1.7
The lifetime τ 0 τ 0 tau_(0)\tau_{0}τ0 is measured in the rest frame of the muon. Due to time dilation, the lifetime in the rest frame of the Earth will be
(3.21) τ = τ 0 γ ( v ) (3.21) τ = τ 0 γ ( v ) {:(3.21)tau=tau_(0)gamma(v):}\begin{equation*} \tau=\tau_{0} \gamma(v) \tag{3.21} \end{equation*}(3.21)τ=τ0γ(v)
where v v vvv is the velocity of the muon relative to Earth. With v = 0.999 c v = 0.999 c v=0.999 cv=0.999 cv=0.999c, one obtains
(3.22) τ = τ 0 1 − v 2 c 2 ≈ 22 τ 0 (3.22) τ = τ 0 1 − v 2 c 2 ≈ 22 τ 0 {:(3.22)tau=(tau_(0))/(sqrt(1-(v^(2))/(c^(2))))~~22tau_(0):}\begin{equation*} \tau=\frac{\tau_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \approx 22 \tau_{0} \tag{3.22} \end{equation*}(3.22)τ=τ01−v2c2≈22τ0
During this time, the muon will move the distance
(3.23) τ v ≈ 22 ⋅ 660 ⋅ 0.999 m ≈ 15 km (3.23) τ v ≈ 22 ⋅ 660 ⋅ 0.999 m ≈ 15 km {:(3.23)tau v~~22*660*0.999m~~15km:}\begin{equation*} \tau v \approx 22 \cdot 660 \cdot 0.999 \mathrm{~m} \approx 15 \mathrm{~km} \tag{3.23} \end{equation*}(3.23)τv≈22⋅660⋅0.999 m≈15 km
in the rest frame of the Earth, which explains that many muons arrive at the surface.
This can also be derived in the muons' rest frame: Due to length contraction, the thickness of the atmosphere ( 10 km ) ( 10 km ) (10km)(10 \mathrm{~km})(10 km) in the rest frame of the Earth will correspond to 10 4 / 22 m ≈ 450 m 10 4 / 22 m ≈ 450 m 10^(4)//22m~~450m10^{4} / 22 \mathrm{~m} \approx 450 \mathrm{~m}104/22 m≈450 m in the rest frame of the muon. During the lifetime of the muon, the Earth will move the distance τ 0 v = 660 ⋅ 0.999 m ≈ 660 m τ 0 v = 660 ⋅ 0.999 m ≈ 660 m tau_(0)v=660*0.999m~~660m\tau_{0} v=660 \cdot 0.999 \mathrm{~m} \approx 660 \mathrm{~m}τ0v=660⋅0.999 m≈660 m in the rest frame of the muon, which is larger than 450 m .
1.8
Introduce the rest frames of the station and the train as K K KKK and K ′ K ′ K^(')K^{\prime}K′, respectively.
a) For the two events that involve the markings on the track, we have
(3.24) Δ t = 0 , Δ x = L a , and Δ x ′ = L 0 (3.24) Δ t = 0 , Δ x = L a ,  and  Δ x ′ = L 0 {:(3.24)Delta t=0","quad Delta x=L_(a)","quad" and "quad Deltax^(')=L_(0):}\begin{equation*} \Delta t=0, \quad \Delta x=L_{a}, \quad \text { and } \quad \Delta x^{\prime}=L_{0} \tag{3.24} \end{equation*}(3.24)Δt=0,Δx=La, and Δx′=L0
When this is inserted into
(3.25) Δ x ′ = γ ( v ) ( Δ x − v Δ t ) (3.25) Δ x ′ = γ ( v ) ( Δ x − v Δ t ) {:(3.25)Deltax^(')=gamma(v)(Delta x-v Delta t):}\begin{equation*} \Delta x^{\prime}=\gamma(v)(\Delta x-v \Delta t) \tag{3.25} \end{equation*}(3.25)Δx′=γ(v)(Δx−vΔt)
we obtain
(3.26) L a = L 0 / γ ( v ) (3.26) L a = L 0 / γ ( v ) {:(3.26)L_(a)=L_(0)//gamma(v):}\begin{equation*} L_{a}=L_{0} / \gamma(v) \tag{3.26} \end{equation*}(3.26)La=L0/γ(v)
b) For the nailings, we have
(3.27) Δ t ′ = 0 , Δ x = L b , and Δ x ′ = L 0 . (3.27) Δ t ′ = 0 , Δ x = L b ,  and  Δ x ′ = L 0 . {:(3.27)Deltat^(')=0","quad Delta x=L_(b)","quad" and "quad Deltax^(')=L_(0).:}\begin{equation*} \Delta t^{\prime}=0, \quad \Delta x=L_{b}, \quad \text { and } \quad \Delta x^{\prime}=L_{0} . \tag{3.27} \end{equation*}(3.27)Δt′=0,Δx=Lb, and Δx′=L0.
Inserting into
(3.28) Δ x = γ ( v ) ( Δ x ′ + v Δ t ′ ) , (3.28) Δ x = γ ( v ) Δ x ′ + v Δ t ′ , {:(3.28)Delta x=gamma(v)(Deltax^(')+v Deltat^('))",":}\begin{equation*} \Delta x=\gamma(v)\left(\Delta x^{\prime}+v \Delta t^{\prime}\right), \tag{3.28} \end{equation*}(3.28)Δx=γ(v)(Δx′+vΔt′),
gives
(3.29) L b = γ ( v ) L 0 . (3.29) L b = γ ( v ) L 0 . {:(3.29)L_(b)=gamma(v)L_(0).:}\begin{equation*} L_{b}=\gamma(v) L_{0} . \tag{3.29} \end{equation*}(3.29)Lb=γ(v)L0.
c) We introduce the events E 1 E 1 E_(1)E_{1}E1 : the end passes B B BBB, and E 2 E 2 E_(2)E_{2}E2 : the front passes A A AAA. For these events, we have
(3.30) Δ t = − L c / c , Δ x = L c , and Δ x ′ = L 0 (3.30) Δ t = − L c / c , Δ x = L c ,  and  Δ x ′ = L 0 {:(3.30)Delta t=-L_(c)//c","quad Delta x=L_(c)","quad" and "quad Deltax^(')=L_(0):}\begin{equation*} \Delta t=-L_{c} / c, \quad \Delta x=L_{c}, \quad \text { and } \quad \Delta x^{\prime}=L_{0} \tag{3.30} \end{equation*}(3.30)Δt=−Lc/c,Δx=Lc, and Δx′=L0
The relation Δ x ′ = γ ( v ) ( Δ x − v Δ t ) Δ x ′ = γ ( v ) ( Δ x − v Δ t ) Deltax^(')=gamma(v)(Delta x-v Delta t)\Delta x^{\prime}=\gamma(v)(\Delta x-v \Delta t)Δx′=γ(v)(Δx−vΔt) gives
(3.31) L c = 1 − v / c 1 + v / c L 0 . (3.31) L c = 1 − v / c 1 + v / c L 0 . {:(3.31)L_(c)=sqrt((1-v//c)/(1+v//c))L_(0).:}\begin{equation*} L_{c}=\sqrt{\frac{1-v / c}{1+v / c}} L_{0} . \tag{3.31} \end{equation*}(3.31)Lc=1−v/c1+v/cL0.
d) We let the events E 1 E 1 E_(1)E_{1}E1 and E 2 E 2 E_(2)E_{2}E2 be the radar pulses reflection against the front and end, respectively. For these events, we have
(3.32) Δ t = ( t 2 − t 1 ) / 2 , Δ t ′ = − L 0 / c , and Δ x ′ = − L 0 . (3.32) Δ t = t 2 − t 1 / 2 , Δ t ′ = − L 0 / c ,  and  Δ x ′ = − L 0 . {:(3.32)Delta t=(t_(2)-t_(1))//2","quad Deltat^(')=-L_(0)//c","quad" and "quad Deltax^(')=-L_(0).:}\begin{equation*} \Delta t=\left(t_{2}-t_{1}\right) / 2, \quad \Delta t^{\prime}=-L_{0} / c, \quad \text { and } \quad \Delta x^{\prime}=-L_{0} . \tag{3.32} \end{equation*}(3.32)Δt=(t2−t1)/2,Δt′=−L0/c, and Δx′=−L0.
Inserting into Δ t = γ ( v ) ( Δ t ′ + v Δ x ′ / c 2 ) Δ t = γ ( v ) Δ t ′ + v Δ x ′ / c 2 Delta t=gamma(v)(Deltat^(')+v Deltax^(')//c^(2))\Delta t=\gamma(v)\left(\Delta t^{\prime}+v \Delta x^{\prime} / c^{2}\right)Δt=γ(v)(Δt′+vΔx′/c2) gives
(3.33) L d = ( t 1 − t 2 ) c 2 = 1 + v / c 1 − v / c L 0 . (3.33) L d = t 1 − t 2 c 2 = 1 + v / c 1 − v / c L 0 . {:(3.33)L_(d)=(t_(1)-t_(2))(c)/(2)=sqrt((1+v//c)/(1-v//c))L_(0).:}\begin{equation*} L_{d}=\left(t_{1}-t_{2}\right) \frac{c}{2}=\sqrt{\frac{1+v / c}{1-v / c}} L_{0} . \tag{3.33} \end{equation*}(3.33)Ld=(t1−t2)c2=1+v/c1−v/cL0.

1.9

Let the coordinates of the front and rear end of the express space cruiser be x F = x F = x_(F)=x_{F}=xF= ( − τ c , L , 0 , 0 ) ( − τ c , L , 0 , 0 ) (-tau c,L,0,0)(-\tau c, L, 0,0)(−τc,L,0,0) and x R = ( 0 , 0 , 0 , 0 ) x R = ( 0 , 0 , 0 , 0 ) x_(R)=(0,0,0,0)x_{R}=(0,0,0,0)xR=(0,0,0,0), respectively, in the rest frame of the cruiser at the time when the rear watchman sees the rear light go on. The difference between these are Δ x = ( − τ c , L , 0 , 0 ) Δ x = ( − τ c , L , 0 , 0 ) Delta x=(-tau c,L,0,0)\Delta x=(-\tau c, L, 0,0)Δx=(−τc,L,0,0). In the rest frame of the hitchhiker, this difference is given by an inverse Lorentz transformation to the rest frame of the asteroid with velocity − v , v − v , v -v,v-v, v−v,v being the velocity of the cruiser in the rest frame of the asteroid. His/her time difference for the lightening of the lanterns is then Δ x ′ 0 / c = − τ cosh ⁡ θ + Δ x ′ 0 / c = − τ cosh ⁡ θ + Deltax^('0)//c=-tau cosh theta+\Delta x^{\prime 0} / c=-\tau \cosh \theta+Δx′0/c=−τcosh⁡θ+ ( L / c ) sinh ⁡ θ ( L / c ) sinh ⁡ θ (L//c)sinh theta(L / c) \sinh \theta(L/c)sinh⁡θ. However, this time difference is equal to zero. Thus, we have
(3.34) τ = L c tanh ⁡ θ = v L c 2 , (3.34) τ = L c tanh ⁡ θ = v L c 2 , {:(3.34)tau=(L)/(c)tanh theta=(vL)/(c^(2))",":}\begin{equation*} \tau=\frac{L}{c} \tanh \theta=\frac{v L}{c^{2}}, \tag{3.34} \end{equation*}(3.34)τ=Lctanh⁡θ=vLc2,
from which we obtain
(3.35) v = τ c 2 L = 1.8 ⋅ 10 5 m / s ≈ 0.0006 c (3.35) v = τ c 2 L = 1.8 ⋅ 10 5 m / s ≈ 0.0006 c {:(3.35)v=(tauc^(2))/(L)=1.8*10^(5)m//s~~0.0006 c:}\begin{equation*} v=\frac{\tau c^{2}}{L}=1.8 \cdot 10^{5} \mathrm{~m} / \mathrm{s} \approx 0.0006 c \tag{3.35} \end{equation*}(3.35)v=τc2L=1.8⋅105 m/s≈0.0006c

1.10

Using the fact that the interval between two points in spacetime is invariant, i.e., Δ s 2 = Δ s ′ 2 Δ s 2 = Δ s ′ 2 Deltas^(2)=Deltas^('2)\Delta s^{2}=\Delta s^{\prime 2}Δs2=Δs′2 or c 2 Δ t 2 − Δ x 2 = c 2 Δ t ′ 2 − Δ x ′ 2 c 2 Δ t 2 − Δ x 2 = c 2 Δ t ′ 2 − Δ x ′ 2 c^(2)Deltat^(2)-Deltax^(2)=c^(2)Deltat^('2)-Deltax^('2)c^{2} \Delta t^{2}-\Delta x^{2}=c^{2} \Delta t^{\prime 2}-\Delta \boldsymbol{x}^{\prime 2}c2Δt2−Δx2=c2Δt′2−Δx′2, gives together with the information in the problem text
(3.36) c 2 ⋅ 0 2 − ℓ 2 = c 2 τ 2 − ℓ ′ 2 (3.36) c 2 ⋅ 0 2 − ℓ 2 = c 2 τ 2 − ℓ ′ 2 {:(3.36)c^(2)*0^(2)-ℓ^(2)=c^(2)tau^(2)-ℓ^('2):}\begin{equation*} c^{2} \cdot 0^{2}-\ell^{2}=c^{2} \tau^{2}-\ell^{\prime 2} \tag{3.36} \end{equation*}(3.36)c2⋅02−ℓ2=c2τ2−ℓ′2
Thus, we find
(3.37) ℓ = ℓ ′ 2 − c 2 τ 2 (3.37) ℓ = ℓ ′ 2 − c 2 τ 2 {:(3.37)ℓ=sqrt(ℓ^('2)-c^(2)tau^(2)):}\begin{equation*} \ell=\sqrt{\ell^{\prime 2}-c^{2} \tau^{2}} \tag{3.37} \end{equation*}(3.37)ℓ=ℓ′2−c2τ2
Now, using the length contraction formula ℓ ′ = ℓ γ ( v ) ℓ ′ = ℓ γ ( v ) ℓ^(')=ℓgamma(v)\ell^{\prime}=\ell \gamma(v)ℓ′=ℓγ(v), where γ ( v ) = 1 1 − v 2 / c 2 γ ( v ) = 1 1 − v 2 / c 2 gamma(v)=(1)/(sqrt(1-v^(2)//c^(2)))\gamma(v)=\frac{1}{\sqrt{1-v^{2} / c^{2}}}γ(v)=11−v2/c2, one obtains
(3.38) ℓ ′ = ℓ ′ 2 − c 2 τ 2 1 1 − v 2 / c 2 (3.38) ℓ ′ = ℓ ′ 2 − c 2 τ 2 1 1 − v 2 / c 2 {:(3.38)ℓ^(')=sqrt(ℓ^('2)-c^(2)tau^(2))(1)/(sqrt(1-v^(2)//c^(2))):}\begin{equation*} \ell^{\prime}=\sqrt{\ell^{\prime 2}-c^{2} \tau^{2}} \frac{1}{\sqrt{1-v^{2} / c^{2}}} \tag{3.38} \end{equation*}(3.38)ℓ′=ℓ′2−c2τ211−v2/c2
from which it follows that
(3.39) v = ± c 2 τ ℓ ′ (3.39) v = ± c 2 τ ℓ ′ {:(3.39)v=+-(c^(2)tau)/(ℓ^(')):}\begin{equation*} v= \pm \frac{c^{2} \tau}{\ell^{\prime}} \tag{3.39} \end{equation*}(3.39)v=±c2τℓ′

1.11

Lorentz contraction only takes place for the projection of the rod that lies along the x x xxx-axis. This projection is ℓ cos ⁡ θ ℓ cos ⁡ θ ℓcos theta\ell \cos \thetaℓcos⁡θ. The orthogonal component is ℓ sin ⁡ θ ℓ sin ⁡ θ ℓsin theta\ell \sin \thetaℓsin⁡θ. Lorentz contraction of the x x xxx-component is then ℓ cos ⁡ θ 1 − v 2 / c 2 ℓ cos ⁡ θ 1 − v 2 / c 2 ℓcos thetasqrt(1-v^(2)//c^(2))\ell \cos \theta \sqrt{1-v^{2} / c^{2}}ℓcos⁡θ1−v2/c2. Therefore, to the moving observer, the length of the rod is
(3.40) ℓ ′ = ( ℓ cos ⁡ θ 1 − v 2 c 2 ) 2 + ( ℓ sin ⁡ θ ) 2 = ℓ 1 − ( v c ) 2 cos 2 ⁡ θ (3.40) ℓ ′ = ℓ cos ⁡ θ 1 − v 2 c 2 2 + ( ℓ sin ⁡ θ ) 2 = ℓ 1 − v c 2 cos 2 ⁡ θ {:(3.40)ℓ^(')=sqrt((ℓcos thetasqrt(1-(v^(2))/(c^(2))))^(2)+(ℓsin theta)^(2))=ℓsqrt(1-((v)/(c))^(2)cos^(2)theta):}\begin{equation*} \ell^{\prime}=\sqrt{\left(\ell \cos \theta \sqrt{1-\frac{v^{2}}{c^{2}}}\right)^{2}+(\ell \sin \theta)^{2}}=\ell \sqrt{1-\left(\frac{v}{c}\right)^{2} \cos ^{2} \theta} \tag{3.40} \end{equation*}(3.40)ℓ′=(ℓcos⁡θ1−v2c2)2+(ℓsin⁡θ)2=ℓ1−(vc)2cos2⁡θ
Thus, the answer is ℓ ′ = ℓ 1 − ( v c ) 2 cos 2 ⁡ θ ℓ ′ = ℓ 1 − v c 2 cos 2 ⁡ θ ℓ^(')=ℓsqrt(1-((v)/(c))^(2)cos^(2)theta)\ell^{\prime}=\ell \sqrt{1-\left(\frac{v}{c}\right)^{2} \cos ^{2} \theta}ℓ′=ℓ1−(vc)2cos2⁡θ.

1.12

The spacetime interval of the two events A A AAA and B B BBB with coordinates x A x A x_(A)x_{A}xA and x B x B x_(B)x_{B}xB is s = ( x A − x B ) 2 s = x A − x B 2 s=(x_(A)-x_(B))^(2)s=\left(x_{A}-x_{B}\right)^{2}s=(xA−xB)2. For observer K K KKK in S S SSS, this is s = − L 2 s = − L 2 s=-L^(2)s=-L^{2}s=−L2. In the rest frame of K ′ K ′ K^(')K^{\prime}K′ we have s = c 2 Δ t ′ 2 − L ′ 2 s = c 2 Δ t ′ 2 − L ′ 2 s=c^(2)Deltat^('2)-L^('2)s=c^{2} \Delta t^{\prime 2}-L^{\prime 2}s=c2Δt′2−L′2. Therefore, we obtain L = L ′ 2 − c 2 Δ t ′ 2 L = L ′ 2 − c 2 Δ t ′ 2 L=sqrt(L^('2)-c^(2)Deltat^('2))L=\sqrt{L^{\prime 2}-c^{2} \Delta t^{\prime 2}}L=L′2−c2Δt′2. Thus, the answer is L = L ′ 2 − c 2 Δ t ′ 2 L = L ′ 2 − c 2 Δ t ′ 2 L=sqrt(L^('2)-c^(2)Deltat^('2))L=\sqrt{L^{\prime 2}-c^{2} \Delta t^{\prime 2}}L=L′2−c2Δt′2.

1.13

The invariant interval is ( x α − x β ) 2 = ( x α ′ − x β ′ ) 2 x α − x β 2 = x α ′ − x β ′ 2 (x_(alpha)-x_(beta))^(2)=(x_(alpha)^(')-x_(beta)^('))^(2)\left(x_{\alpha}-x_{\beta}\right)^{2}=\left(x_{\alpha}^{\prime}-x_{\beta}^{\prime}\right)^{2}(xα−xβ)2=(xα′−xβ′)2.
a) Denote the distance to the β β beta\betaβ event for S ′ S ′ S^(')S^{\prime}S′ by L ′ L ′ L^(')L^{\prime}L′. We set c = 1 c = 1 c=1c=1c=1 and calculate the length in ly. Then, we have
(3.41) 2 2 1 y 2 − 10 2 ly 2 = 1 2 ly 2 − L ′ 2 (3.41) 2 2 1 y 2 − 10 2 ly 2 = 1 2 ly 2 − L ′ 2 {:(3.41)2^(2)1y^(2)-10^(2)ly^(2)=1^(2)ly^(2)-L^('2):}\begin{equation*} 2^{2} 1 y^{2}-10^{2} \mathrm{ly}^{2}=1^{2} \mathrm{ly}^{2}-L^{\prime 2} \tag{3.41} \end{equation*}(3.41)221y2−102ly2=12ly2−L′2
Solving this equation gives
(3.42) L ′ = 100 + 1 − 4 ly = 97 ly ≈ 9.85 ly (3.42) L ′ = 100 + 1 − 4 ly = 97 ly ≈ 9.85 ly {:(3.42)L^(')=sqrt(100+1-4)ly=sqrt97ly~~9.85ly:}\begin{equation*} L^{\prime}=\sqrt{100+1-4} \mathrm{ly}=\sqrt{97} \mathrm{ly} \approx 9.85 \mathrm{ly} \tag{3.42} \end{equation*}(3.42)L′=100+1−4ly=97ly≈9.85ly
b) The Lorentz transformation from K K KKK to K ′ K ′ K^(')K^{\prime}K′ gives
(3.43) 1 = 2 γ − 10 v γ (3.44) L ′ / ly = − 2 v γ + 10 γ (3.43) 1 = 2 γ − 10 v γ (3.44) L ′ / ly = − 2 v γ + 10 γ {:[(3.43)1=2gamma-10 v gamma],[(3.44)L^(')//ly=-2v gamma+10 gamma]:}\begin{align*} 1 & =2 \gamma-10 v \gamma \tag{3.43}\\ L^{\prime} / \mathrm{ly} & =-2 v \gamma+10 \gamma \tag{3.44} \end{align*}(3.43)1=2γ−10vγ(3.44)L′/ly=−2vγ+10γ
where γ = 1 / 1 − v 2 γ = 1 / 1 − v 2 gamma=1//sqrt(1-v^(2))\gamma=1 / \sqrt{1-v^{2}}γ=1/1−v2. Solving this equation yields v ≈ 0.1 v ≈ 0.1 v~~0.1v \approx 0.1v≈0.1 as the permissible root.
1.14
The smallest energy required for a muon to hit the ground within its mean life in the rest frame of the Earth is given when the direction of the muon is vertical. In this case, the muon must travel the distance ℓ = 10 km ℓ = 10 km ℓ=10km\ell=10 \mathrm{~km}ℓ=10 km in the time τ τ tau\tauτ, which is the time dilated mean life τ 0 τ 0 tau_(0)\tau_{0}τ0 of the muon in its rest frame. It follows that
(3.45) ℓ ≤ v τ = v γ ( v ) τ 0 (3.45) ℓ ≤ v τ = v γ ( v ) τ 0 {:(3.45)ℓ <= v tau=v gamma(v)tau_(0):}\begin{equation*} \ell \leq v \tau=v \gamma(v) \tau_{0} \tag{3.45} \end{equation*}(3.45)ℓ≤vτ=vγ(v)τ0
The velocity of the muon is given by v = p c 2 / E v = p c 2 / E v=pc^(2)//Ev=p c^{2} / Ev=pc2/E, and thus, we have
(3.46) γ ( v ) = 1 1 − ( v / c ) 2 = E E 2 − p 2 c 2 = E m c 2 (3.46) γ ( v ) = 1 1 − ( v / c ) 2 = E E 2 − p 2 c 2 = E m c 2 {:(3.46)gamma(v)=(1)/(sqrt(1-(v//c)^(2)))=(E)/(sqrt(E^(2)-p^(2)c^(2)))=(E)/(mc^(2)):}\begin{equation*} \gamma(v)=\frac{1}{\sqrt{1-(v / c)^{2}}}=\frac{E}{\sqrt{E^{2}-p^{2} c^{2}}}=\frac{E}{m c^{2}} \tag{3.46} \end{equation*}(3.46)γ(v)=11−(v/c)2=EE2−p2c2=Emc2
Inserting this into the above inequality, we obtain
(3.47) ℓ ≤ p c m c 2 c τ 0 (3.47) ℓ ≤ p c m c 2 c τ 0 {:(3.47)ℓ <= (pc)/(mc^(2))ctau_(0):}\begin{equation*} \ell \leq \frac{p c}{m c^{2}} c \tau_{0} \tag{3.47} \end{equation*}(3.47)ℓ≤pcmc2cτ0
Inserting the numerical values, we find that p c ≳ 15 m c 2 ≃ 1.6 GeV p c ≳ 15 m c 2 ≃ 1.6 GeV pc≳15 mc^(2)≃1.6GeVp c \gtrsim 15 m c^{2} \simeq 1.6 \mathrm{GeV}pc≳15mc2≃1.6GeV, and thus, it holds that E ≃ p c ≳ 1.6 GeV E ≃ p c ≳ 1.6 GeV E≃pc≳1.6GeVE \simeq p c \gtrsim 1.6 \mathrm{GeV}E≃pc≳1.6GeV.

1.15

A time interval for a muon and a time interval for an observer in the lab frame are related through the time dilation formula
(3.48) γ ( v ) d τ = d t (3.48) γ ( v ) d τ = d t {:(3.48)gamma(v)d tau=dt:}\begin{equation*} \gamma(v) d \tau=d t \tag{3.48} \end{equation*}(3.48)γ(v)dτ=dt
where d τ d τ d taud \taudτ is the time interval for the muon, d t d t dtd tdt is the time interval for the observer in the lab frame, and v v vvv is the muon velocity. The muon velocity is constant (since the total energy is constant) and given by
(3.49) v = p E = E 2 − m 2 E = 1 − m 2 E 2 (3.49) v = p E = E 2 − m 2 E = 1 − m 2 E 2 {:(3.49)v=(p)/(E)=(sqrt(E^(2)-m^(2)))/(E)=sqrt(1-(m^(2))/(E^(2))):}\begin{equation*} v=\frac{p}{E}=\frac{\sqrt{E^{2}-m^{2}}}{E}=\sqrt{1-\frac{m^{2}}{E^{2}}} \tag{3.49} \end{equation*}(3.49)v=pE=E2−m2E=1−m2E2
where p p ppp is the muon momentum, E E EEE is the muon energy, and m m mmm is the muon mass. It follows that
(3.50) γ ( v ) = E m (3.50) γ ( v ) = E m {:(3.50)gamma(v)=(E)/(m):}\begin{equation*} \gamma(v)=\frac{E}{m} \tag{3.50} \end{equation*}(3.50)γ(v)=Em
and thus, we obtain
(3.51) t = E m τ = 1 GeV 106 MeV τ ≃ 10 τ (3.51) t = E m τ = 1 GeV 106 MeV τ ≃ 10 τ {:(3.51)t=(E)/(m)tau=(1GeV)/(106MeV)tau≃10 tau:}\begin{equation*} t=\frac{E}{m} \tau=\frac{1 \mathrm{GeV}}{106 \mathrm{MeV}} \tau \simeq 10 \tau \tag{3.51} \end{equation*}(3.51)t=Emτ=1GeV106MeVτ≃10τ
The average lifetime of the muon in the lab frame is therefore 22 μ s 22 μ s 22 mus22 \mu \mathrm{~s}22μ s. Since the muon is highly relativistic ( E ≫ m ) ( E ≫ m ) (E≫m)(E \gg m)(E≫m), the length traveled by the muon in the lab frame in the average lifetime is given by
(3.52) ℓ = v t ≃ c t ≃ 3 ⋅ 10 8 m / s ⋅ 22 ⋅ 10 − 6 s = 6600 m . (3.52) ℓ = v t ≃ c t ≃ 3 ⋅ 10 8 m / s ⋅ 22 ⋅ 10 − 6 s = 6600 m . {:(3.52)ℓ=vt≃ct≃3*10^(8)m//s*22*10^(-6)s=6600m.:}\begin{equation*} \ell=v t \simeq c t \simeq 3 \cdot 10^{8} \mathrm{~m} / \mathrm{s} \cdot 22 \cdot 10^{-6} \mathrm{~s}=6600 \mathrm{~m} . \tag{3.52} \end{equation*}(3.52)ℓ=vt≃ct≃3⋅108 m/s⋅22⋅10−6 s=6600 m.
The circumference of the circular accelerator is given by
(3.53) L = 2 π r ≃ 300 m (3.53) L = 2 π r ≃ 300 m {:(3.53)L=2pi r≃300m:}\begin{equation*} L=2 \pi r \simeq 300 \mathrm{~m} \tag{3.53} \end{equation*}(3.53)L=2πr≃300 m
Thus, the average number of turns taken by a muon is given by
(3.54) N = ℓ L ≃ 22 (3.54) N = ℓ L ≃ 22 {:(3.54)N=(ℓ)/(L)≃22:}\begin{equation*} N=\frac{\ell}{L} \simeq 22 \tag{3.54} \end{equation*}(3.54)N=ℓL≃22

1.16

We let ℓ = 1 ℓ = 1 ℓ=1\ell=1ℓ=1 to simplify the notation.
a) We use the standard formulas for Lorentz contractions, and thus, we have
(3.55) a ′ = a / γ = 3 / γ , b ′ = b = 4 , c ′ = ( a ′ ) 2 + ( b ′ ) 2 = 9 / γ 2 + 16 (3.55) a ′ = a / γ = 3 / γ , b ′ = b = 4 , c ′ = a ′ 2 + b ′ 2 = 9 / γ 2 + 16 {:(3.55)a^(')=a//gamma=3//gamma","quadb^(')=b=4","quadc^(')=sqrt((a^('))^(2)+(b^('))^(2))=sqrt(9//gamma^(2)+16):}\begin{equation*} a^{\prime}=a / \gamma=3 / \gamma, \quad b^{\prime}=b=4, \quad c^{\prime}=\sqrt{\left(a^{\prime}\right)^{2}+\left(b^{\prime}\right)^{2}}=\sqrt{9 / \gamma^{2}+16} \tag{3.55} \end{equation*}(3.55)a′=a/γ=3/γ,b′=b=4,c′=(a′)2+(b′)2=9/γ2+16
where γ = 1 / 1 − ( v / c ) 2 γ = 1 / 1 − ( v / c ) 2 gamma=1//sqrt(1-(v//c)^(2))\gamma=1 / \sqrt{1-(v / c)^{2}}γ=1/1−(v/c)2, and thus, the area becomes
(3.56) A ′ = a ′ b ′ / 2 = 6 / γ = A / γ (3.56) A ′ = a ′ b ′ / 2 = 6 / γ = A / γ {:(3.56)A^(')=a^(')b^(')//2=6//gamma=A//gamma:}\begin{equation*} A^{\prime}=a^{\prime} b^{\prime} / 2=6 / \gamma=A / \gamma \tag{3.56} \end{equation*}(3.56)A′=a′b′/2=6/γ=A/γ
where A = 6 A = 6 A=6A=6A=6 is the area in K K KKK.
b) Let h h hhh be the height of the triangle in K ′ K ′ K^(')K^{\prime}K′. Since A = 6 = c h / 2 A = 6 = c h / 2 A=6=ch//2A=6=c h / 2A=6=ch/2, we get h = h = h=h=h= 12 / 5 12 / 5 12//512 / 512/5. We have c ′ = c / γ , h ′ = h c ′ = c / γ , h ′ = h c^(')=c//gamma,h^(')=hc^{\prime}=c / \gamma, h^{\prime}=hc′=c/γ,h′=h, and thus, the area becomes A ′ = c ′ h ′ / 2 = A / γ A ′ = c ′ h ′ / 2 = A / γ A^(')=c^(')h^(')//2=A//gammaA^{\prime}=c^{\prime} h^{\prime} / 2=A / \gammaA′=c′h′/2=A/γ as in a). Let c 1 c 1 c_(1)c_{1}c1 be the distance in K K KKK from the corner, where the a a aaa - and c c ccc-sides of the triangle meet in the point, where the height intersects the c c ccc-side in a right angle. Obviously, c 1 = a 2 − h 2 c 1 = a 2 − h 2 c_(1)=sqrt(a^(2)-h^(2))c_{1}=\sqrt{a^{2}-h^{2}}c1=a2−h2. In K ′ K ′ K^(')K^{\prime}K′, the same distance is c 1 ′ = c 1 / γ c 1 ′ = c 1 / γ c_(1)^(')=c_(1)//gammac_{1}^{\prime}=c_{1} / \gammac1′=c1/γ, and thus, we obtain
(3.57) a ′ = ( c 1 ′ ) 2 + ( h ′ ) 2 = ( a 2 − h 2 ) / γ 2 + h 2 = 9 / γ 2 + ( 144 / 25 ) ( 1 − 1 / γ 2 ) (3.57) a ′ = c 1 ′ 2 + h ′ 2 = a 2 − h 2 / γ 2 + h 2 = 9 / γ 2 + ( 144 / 25 ) 1 − 1 / γ 2 {:(3.57)a^(')=sqrt((c_(1)^('))^(2)+(h^('))^(2))=sqrt((a^(2)-h^(2))//gamma^(2)+h^(2))=sqrt(9//gamma^(2)+(144//25)(1-1//gamma^(2))):}\begin{equation*} a^{\prime}=\sqrt{\left(c_{1}^{\prime}\right)^{2}+\left(h^{\prime}\right)^{2}}=\sqrt{\left(a^{2}-h^{2}\right) / \gamma^{2}+h^{2}}=\sqrt{9 / \gamma^{2}+(144 / 25)\left(1-1 / \gamma^{2}\right)} \tag{3.57} \end{equation*}(3.57)a′=(c1′)2+(h′)2=(a2−h2)/γ2+h2=9/γ2+(144/25)(1−1/γ2)
and similarly, we find that
(3.58) b ′ = 16 / γ 2 + ( 144 / 25 ) ( 1 − 1 / γ 2 ) (3.58) b ′ = 16 / γ 2 + ( 144 / 25 ) 1 − 1 / γ 2 {:(3.58)b^(')=sqrt(16//gamma^(2)+(144//25)(1-1//gamma^(2))):}\begin{equation*} b^{\prime}=\sqrt{16 / \gamma^{2}+(144 / 25)\left(1-1 / \gamma^{2}\right)} \tag{3.58} \end{equation*}(3.58)b′=16/γ2+(144/25)(1−1/γ2)
1.17
The rest frame of the pole is denoted S S SSS with coordinates X = ( t , x , y , z ) X = ( t , x , y , z ) X=(t,x,y,z)X=(t, x, y, z)X=(t,x,y,z), where the pole has length L L LLL. The lab frame is denoted S ′ S ′ S^(')S^{\prime}S′ with coordinates X ′ = ( t ′ , x ′ , y ′ , z ′ ) X ′ = t ′ , x ′ , y ′ , z ′ X^(')=(t^('),x^('),y^('),z^('))X^{\prime}=\left(t^{\prime}, x^{\prime}, y^{\prime}, z^{\prime}\right)X′=(t′,x′,y′,z′), in which the pole length is L ′ = L / γ L ′ = L / γ L^(')=L//gammaL^{\prime}=L / \gammaL′=L/γ, where γ = 1 / 1 − v 2 / c 2 γ = 1 / 1 − v 2 / c 2 gamma=1//sqrt(1-v^(2)//c^(2))\gamma=1 / \sqrt{1-v^{2} / c^{2}}γ=1/1−v2/c2. In the lab frame, the pole is moving in the negative x x xxx-direction with speed v v vvv, parallel to the x x xxx-axis at some distance y y yyy. Everything occurs in the spatial plane z = 0 z = 0 z=0z=0z=0, so we can ignore the z z zzz-direction. Denote the events at which light is emitted from front of the pole by A, from the mark on the pole by B, and from the end of the pole by C. The three rays of light all reach the origin event in S ′ S ′ S^(')S^{\prime}S′, i.e., the spatial origin at t ′ = 0 t ′ = 0 t^(')=0t^{\prime}=0t′=0. Since the three light rays are emitted from different positions, but all reach the origin at the same time, the light rays were emitted at different times t ′ t ′ t^(')t^{\prime}t′. We now consider each of the three light rays separately.
Event A: Let X A ′ = ( t ′ , x ′ , y ′ ) X A ′ = t ′ , x ′ , y ′ X_(A)^(')=(t^('),x^('),y^('))X_{\mathrm{A}}^{\prime}=\left(t^{\prime}, x^{\prime}, y^{\prime}\right)XA′=(t′,x′,y′) denote the coordinates in S ′ S ′ S^(')S^{\prime}S′ of the event A. Since the line element along the 4-path of a light signal vanishes, we have
(3.59) t ′ 2 = x ′ 2 + y ′ 2 (3.59) t ′ 2 = x ′ 2 + y ′ 2 {:(3.59)t^('2)=x^('2)+y^('2):}\begin{equation*} t^{\prime 2}=x^{\prime 2}+y^{\prime 2} \tag{3.59} \end{equation*}(3.59)t′2=x′2+y′2
The light ray defines an angle π / 3 π / 3 pi//3\pi / 3π/3 with the x x xxx-axis, and therefore, we find
(3.60) y ′ x ′ = tan ⁡ π 3 = 3 (3.60) y ′ x ′ = tan ⁡ π 3 = 3 {:(3.60)(y^('))/(x^('))=tan((pi)/(3))=sqrt3:}\begin{equation*} \frac{y^{\prime}}{x^{\prime}}=\tan \frac{\pi}{3}=\sqrt{3} \tag{3.60} \end{equation*}(3.60)y′x′=tan⁡π3=3
Combining the two equations, we get
(3.61) t ′ 2 = 4 x ′ 2 , y ′ = 3 x ′ (3.61) t ′ 2 = 4 x ′ 2 , y ′ = 3 x ′ {:(3.61)t^('2)=4x^('2)","quady^(')=sqrt3x^('):}\begin{equation*} t^{\prime 2}=4 x^{\prime 2}, \quad y^{\prime}=\sqrt{3} x^{\prime} \tag{3.61} \end{equation*}(3.61)t′2=4x′2,y′=3x′
Defining the time at which the event A occurs to be t ′ = − δ A t ′ = − δ A t^(')=-delta_(A)t^{\prime}=-\delta_{\mathrm{A}}t′=−δA, we obtain the coordinates of A in S ′ S ′ S^(')S^{\prime}S′ as
(3.62) X A ′ = ( − 1 , 1 2 , 3 2 ) δ A (3.62) X A ′ = − 1 , 1 2 , 3 2 δ A {:(3.62)X_(A)^(')=(-1,(1)/(2),(sqrt3)/(2))delta_(A):}\begin{equation*} X_{\mathrm{A}}^{\prime}=\left(-1, \frac{1}{2}, \frac{\sqrt{3}}{2}\right) \delta_{\mathrm{A}} \tag{3.62} \end{equation*}(3.62)XA′=(−1,12,32)δA
Event B: Let X B ′ = ( t ′ , x ′ , y ′ ) X B ′ = t ′ , x ′ , y ′ X_(B)^(')=(t^('),x^('),y^('))X_{\mathrm{B}}^{\prime}=\left(t^{\prime}, x^{\prime}, y^{\prime}\right)XB′=(t′,x′,y′) denote the coordinates in S ′ S ′ S^(')S^{\prime}S′ of the event B. In this case, the angle is π / 4 π / 4 pi//4\pi / 4π/4, so we have
(3.63) t ′ 2 = x ′ 2 + y ′ 2 , y ′ x ′ = tan ⁡ π 4 = 1 (3.63) t ′ 2 = x ′ 2 + y ′ 2 , y ′ x ′ = tan ⁡ π 4 = 1 {:(3.63)t^('2)=x^('2)+y^('2)","quad(y^('))/(x^('))=tan((pi)/(4))=1:}\begin{equation*} t^{\prime 2}=x^{\prime 2}+y^{\prime 2}, \quad \frac{y^{\prime}}{x^{\prime}}=\tan \frac{\pi}{4}=1 \tag{3.63} \end{equation*}(3.63)t′2=x′2+y′2,y′x′=tan⁡π4=1
which means that
(3.64) t ′ 2 = 2 x ′ 2 , y ′ = x ′ (3.64) t ′ 2 = 2 x ′ 2 , y ′ = x ′ {:(3.64)t^('2)=2x^('2)","quady^(')=x^('):}\begin{equation*} t^{\prime 2}=2 x^{\prime 2}, \quad y^{\prime}=x^{\prime} \tag{3.64} \end{equation*}(3.64)t′2=2x′2,y′=x′
Thus, defining the time of the event B as t ′ = − δ B t ′ = − δ B t^(')=-delta_(B)t^{\prime}=-\delta_{\mathrm{B}}t′=−δB, we obtain
(3.65) X B ′ = ( − 1 , 1 2 , 1 2 ) δ B . (3.65) X B ′ = − 1 , 1 2 , 1 2 δ B . {:(3.65)X_(B)^(')=(-1,(1)/(sqrt2),(1)/(sqrt2))delta_(B).:}\begin{equation*} X_{\mathrm{B}}^{\prime}=\left(-1, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \delta_{\mathrm{B}} . \tag{3.65} \end{equation*}(3.65)XB′=(−1,12,12)δB.
Event C: Let X C ′ = ( t ′ , x ′ , y ′ ) X C ′ = t ′ , x ′ , y ′ X_(C)^(')=(t^('),x^('),y^('))X_{\mathrm{C}}^{\prime}=\left(t^{\prime}, x^{\prime}, y^{\prime}\right)XC′=(t′,x′,y′) denote the coordinates in S ′ S ′ S^(')S^{\prime}S′ of the event C. Similar to events A and B , defining the time of the event C as t ′ = − δ C t ′ = − δ C t^(')=-delta_(C)t^{\prime}=-\delta_{\mathrm{C}}t′=−δC, we obtain
(3.66) X C ′ = ( − 1 , 3 2 , 1 2 ) δ C (3.66) X C ′ = − 1 , 3 2 , 1 2 δ C {:(3.66)X_(C)^(')=(-1,(sqrt3)/(2),(1)/(2))delta_(C):}\begin{equation*} X_{\mathrm{C}}^{\prime}=\left(-1, \frac{\sqrt{3}}{2}, \frac{1}{2}\right) \delta_{\mathrm{C}} \tag{3.66} \end{equation*}(3.66)XC′=(−1,32,12)δC
Since the values of y ′ y ′ y^(')y^{\prime}y′ are all equal for the three events A, B, and C, we find that
(3.67) 3 2 δ A = 1 2 δ B = 1 2 δ C , (3.67) 3 2 δ A = 1 2 δ B = 1 2 δ C , {:(3.67)(sqrt3)/(2)delta_(A)=(1)/(sqrt2)delta_(B)=(1)/(2)delta_(C)",":}\begin{equation*} \frac{\sqrt{3}}{2} \delta_{\mathrm{A}}=\frac{1}{\sqrt{2}} \delta_{\mathrm{B}}=\frac{1}{2} \delta_{\mathrm{C}}, \tag{3.67} \end{equation*}(3.67)32δA=12δB=12δC,
which means that the coordinates of all three events can be expressed in the time δ A δ A delta_(A)\delta_{\mathrm{A}}δA only, i.e.,
(3.68) X A ′ = ( − 1 , 1 2 , 3 2 ) δ A , X B ′ ( − 3 2 , 3 2 , 3 2 ) δ A , X C ′ = ( − 3 , 3 2 , 3 2 ) δ A . (3.68) X A ′ = − 1 , 1 2 , 3 2 δ A , X B ′ − 3 2 , 3 2 , 3 2 δ A , X C ′ = − 3 , 3 2 , 3 2 δ A . {:(3.68)X_(A)^(')=(-1,(1)/(2),(sqrt3)/(2))delta_(A)","quadX_(B)^(')(-sqrt((3)/(2)),(sqrt3)/(2),(sqrt3)/(2))delta_(A)","quadX_(C)^(')=(-sqrt3,(3)/(2),(sqrt3)/(2))delta_(A).:}\begin{equation*} X_{\mathrm{A}}^{\prime}=\left(-1, \frac{1}{2}, \frac{\sqrt{3}}{2}\right) \delta_{\mathrm{A}}, \quad X_{\mathrm{B}}^{\prime}\left(-\sqrt{\frac{3}{2}}, \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}\right) \delta_{\mathrm{A}}, \quad X_{\mathrm{C}}^{\prime}=\left(-\sqrt{3}, \frac{3}{2}, \frac{\sqrt{3}}{2}\right) \delta_{\mathrm{A}} . \tag{3.68} \end{equation*}(3.68)XA′=(−1,12,32)δA,XB′(−32,32,32)δA,XC′=(−3,32,32)δA.
The event A occurs at the front of the pole at x ′ = δ A / 2 x ′ = δ A / 2 x^(')=delta_(A)//2x^{\prime}=\delta_{\mathrm{A}} / 2x′=δA/2 at time − δ A − δ A -delta_(A)-\delta_{\mathrm{A}}−δA, whereas the event C occurs at the end of the pole at x ′ = 3 δ A / 2 x ′ = 3 δ A / 2 x^(')=3delta_(A)//2x^{\prime}=3 \delta_{\mathrm{A}} / 2x′=3δA/2 at time − 3 δ A − 3 δ A -sqrt3delta_(A)-\sqrt{3} \delta_{\mathrm{A}}−3δA. The pole moves at speed v v vvv in the negative x x xxx-direction. Therefore, at time − δ A − δ A -delta_(A)-\delta_{\mathrm{A}}−δA, the end of the pole is located at
(3.69) x C ′ − v ( t A ′ − t C ′ ) = [ 3 2 − v ( 3 − 1 ) ] δ A . (3.69) x C ′ − v t A ′ − t C ′ = 3 2 − v ( 3 − 1 ) δ A . {:(3.69)x_(C)^(')-v(t_(A)^(')-t_(C)^('))=[(3)/(2)-v(sqrt3-1)]delta_(A).:}\begin{equation*} x_{\mathrm{C}}^{\prime}-v\left(t_{\mathrm{A}}^{\prime}-t_{\mathrm{C}}^{\prime}\right)=\left[\frac{3}{2}-v(\sqrt{3}-1)\right] \delta_{\mathrm{A}} . \tag{3.69} \end{equation*}(3.69)xC′−v(tA′−tC′)=[32−v(3−1)]δA.
In S ′ S ′ S^(')S^{\prime}S′, we find that the length of the pole is the difference between this and x A ′ = δ A / 2 x A ′ = δ A / 2 x_(A)^(')=delta_(A)//2x_{\mathrm{A}}^{\prime}=\delta_{\mathrm{A}} / 2xA′=δA/2, which occurs simultaneously at the front, so
(3.70) L ′ = [ 1 − v ( 3 − 1 ) ] δ A (3.70) L ′ = [ 1 − v ( 3 − 1 ) ] δ A {:(3.70)L^(')=[1-v(sqrt3-1)]delta_(A):}\begin{equation*} L^{\prime}=[1-v(\sqrt{3}-1)] \delta_{\mathrm{A}} \tag{3.70} \end{equation*}(3.70)L′=[1−v(3−1)]δA
However, in S ′ S ′ S^(')S^{\prime}S′, we know that the length of the pole is L ′ = L / γ L ′ = L / γ L^(')=L//gammaL^{\prime}=L / \gammaL′=L/γ, which means that we have determined δ A δ A delta_(A)\delta_{\mathrm{A}}δA to be
(3.71) δ A = L ′ 1 − v ( 3 − 1 ) = L γ [ 1 − v ( 3 − 1 ) ] (3.71) δ A = L ′ 1 − v ( 3 − 1 ) = L γ [ 1 − v ( 3 − 1 ) ] {:(3.71)delta_(A)=(L^('))/(1-v(sqrt3-1))=(L)/(gamma[1-v(sqrt3-1)]):}\begin{equation*} \delta_{\mathrm{A}}=\frac{L^{\prime}}{1-v(\sqrt{3}-1)}=\frac{L}{\gamma[1-v(\sqrt{3}-1)]} \tag{3.71} \end{equation*}(3.71)δA=L′1−v(3−1)=Lγ[1−v(3−1)]
and therefore, we have determined all coordinates of the events A , B A , B A,B\mathrm{A}, \mathrm{B}A,B, and C in S ′ S ′ S^(')S^{\prime}S′.
Now, we determine the distance between the front of the pole and the mark on the pole. At time − δ A − δ A -delta_(A)-\delta_{\mathrm{A}}−δA, the x ′ x ′ x^(')x^{\prime}x′-coordinate of the mark is given by
(3.72) x B ′ − v ( t A ′ − t B ′ ) = [ 3 2 − v ( 3 2 − 1 ) ] δ A . (3.72) x B ′ − v t A ′ − t B ′ = 3 2 − v 3 2 − 1 δ A . {:(3.72)x_(B)^(')-v(t_(A)^(')-t_(B)^('))=[(sqrt3)/(2)-v(sqrt((3)/(2))-1)]delta_(A).:}\begin{equation*} x_{\mathrm{B}}^{\prime}-v\left(t_{\mathrm{A}}^{\prime}-t_{\mathrm{B}}^{\prime}\right)=\left[\frac{\sqrt{3}}{2}-v\left(\sqrt{\frac{3}{2}}-1\right)\right] \delta_{\mathrm{A}} . \tag{3.72} \end{equation*}(3.72)xB′−v(tA′−tB′)=[32−v(32−1)]δA.
Thus, the distance between the front and the mark is
(3.73) L ~ ′ = [ 3 2 − 1 2 − v ( 3 2 − 1 ) ] δ A (3.73) L ~ ′ = 3 2 − 1 2 − v 3 2 − 1 δ A {:(3.73) tilde(L)^(')=[(sqrt3)/(2)-(1)/(2)-v(sqrt((3)/(2))-1)]delta_(A):}\begin{equation*} \tilde{L}^{\prime}=\left[\frac{\sqrt{3}}{2}-\frac{1}{2}-v\left(\sqrt{\frac{3}{2}}-1\right)\right] \delta_{\mathrm{A}} \tag{3.73} \end{equation*}(3.73)L~′=[32−12−v(32−1)]δA
Therefore, in S ′ S ′ S^(')S^{\prime}S′, the ratio r r rrr between the distance from the front of the pole to the mark on the pole and the length of the whole pole is
(3.74) r ≡ L ~ ′ L ′ = 3 2 − 1 2 − v ( 3 2 − 1 ) 1 − v ( 3 − 1 ) ≈ 0.37 − 0.22 v 1 − 0.73 v (3.74) r ≡ L ~ ′ L ′ = 3 2 − 1 2 − v 3 2 − 1 1 − v ( 3 − 1 ) ≈ 0.37 − 0.22 v 1 − 0.73 v {:(3.74)r-=( tilde(L)^('))/(L^('))=((sqrt3)/(2)-(1)/(2)-v(sqrt((3)/(2))-1))/(1-v(sqrt3-1))~~0.37-(0.22 v)/(1-0.73 v):}\begin{equation*} r \equiv \frac{\tilde{L}^{\prime}}{L^{\prime}}=\frac{\frac{\sqrt{3}}{2}-\frac{1}{2}-v\left(\sqrt{\frac{3}{2}}-1\right)}{1-v(\sqrt{3}-1)} \approx 0.37-\frac{0.22 v}{1-0.73 v} \tag{3.74} \end{equation*}(3.74)r≡L~′L′=32−12−v(32−1)1−v(3−1)≈0.37−0.22v1−0.73v
Note that r r rrr is the same in all inertial frames due to the linearity of Lorentz transformations. For example, in the rest frame of a pole with length L L LLL and the front at the origin, the mark would be located at the x x xxx-coordinate r L r L rLr LrL.

1.18

a) In the initial rest frame of the spaceships, they are at a constant distance L L LLL from each other during the entire accelerating phase. However, this length is the length contraction of the string in the new rest frame and the proper length will then be L ′ = γ L L ′ = γ L L^(')=gamma LL^{\prime}=\gamma LL′=γL. Trying to force the string to be of length L L LLL in the initial rest frame will therefore break it.
b) The initial distance between the two spaceships is 40 km . The spaceships accelerate with a constant acceleration a = 1 / 50 c / s a = 1 / 50 c / s a=1//50c//sa=1 / 50 \mathrm{c} / \mathrm{s}a=1/50c/s. When the spaceships stop after 30 s in the initial rest frame, the speed of the spaceships will then be v = a t = 3 c / 5 v = a t = 3 c / 5 v=at=3c//5v=a t=3 c / 5v=at=3c/5. Hence, the gamma factor can be computed to be
(3.75) γ = 1 1 − v 2 c 2 = 1 1 − ( 3 5 ) 2 = 5 4 (3.75) γ = 1 1 − v 2 c 2 = 1 1 − 3 5 2 = 5 4 {:(3.75)gamma=(1)/(sqrt(1-(v^(2))/(c^(2))))=(1)/(sqrt(1-((3)/(5))^(2)))=(5)/(4):}\begin{equation*} \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\frac{1}{\sqrt{1-\left(\frac{3}{5}\right)^{2}}}=\frac{5}{4} \tag{3.75} \end{equation*}(3.75)γ=11−v2c2=11−(35)2=54
Thus, the distance between the two spaceships is L ′ = γ L = 5 / 4 ⋅ 40 km = 50 km L ′ = γ L = 5 / 4 ⋅ 40 km = 50 km L^(')=gamma L=5//4*40km=50kmL^{\prime}=\gamma L=5 / 4 \cdot 40 \mathrm{~km}=50 \mathrm{~km}L′=γL=5/4⋅40 km=50 km.

1.19

Spacetime diagrams showing the events described in the problem are shown in Figure 3.2. Part (a) of the diagram depicts the events using the time and x x xxx axes of the rest frame of Professor Einstein as orthogonal, while part (b) is the same spacetime diagram with the axes of Professor Wolf orthogonal. The coordinate axes of Professor Einstein are marked by t E t E t_(E)t_{\mathrm{E}}tE and x E x E x_(E)x_{\mathrm{E}}xE, respectively, while those of Professor Wolf are marked by t W t W t_(W)t_{\mathrm{W}}tW and x W x W x_(W)x_{\mathrm{W}}xW, respectively. The light gray shaded region is the train, the thick black lines and the thick dark gray lines represent the worldlines of Professor Einstein and Professor Wolf, respectively, while the thick white lines represent the light signals.
Figure 3.2 Spacetime diagrams describing the events of Problem 1.19.
From the diagrams it is clear that
a) The light signals will always reach Professor Einstein at the same time. This is something physically observable and cannot depend on the set of coordinates used to describe the events. (In fact, a light signal reaching Professor Einstein is an event and both light signals reach him at the same event).
b) Professor Wolf sees the light signals being reflected at different times. This can be seen as the line between the reflection points is not parallel to the spatial axis in Professor Wolf's rest frame, i.e., it is not a simultaneity in that frame.
c) The signals will reach Professor Wolf at different times. The signal from the back of the train has already reached Professor Wolf before it reaches Professor Einstein, while the signal from the front reflection will reach him after it reaches Professor Einstein.

1.20

Let us consider the situation seen in the frame S S SSS where the two spaceships have equal but opposite velocities. The situation can then be described through the spacetime diagram in Figure 3.3. Since the velocities of both spaceships relative to this frame are the same, they are equally Lorentz contracted and the ship with rest length 2 L 2 L 2L2 L2L is twice as long (i.e., light gray shaded region) as the ship with rest length L L LLL (i.e., dark gray shaded region). The coordinates have been chosen such that A A AAA occurs in the spacetime origin and the event B B BBB has been marked in the diagram. The time axes of both the frame S ′ S ′ S^(')S^{\prime}S′, in which the shorter ship is at rest, and S ′ ′ S ′ ′ S^('')S^{\prime \prime}S′′, in which the longer is at rest, are also shown. The dark (light) gray line represents the surface simultaneous with B B BBB in S ′ ( S ′ ′ ) S ′ S ′ ′ S^(')(S^(''))S^{\prime}\left(S^{\prime \prime}\right)S′(S′′) and the intersection with the t ′ ( t ′ ′ ) t ′ t ′ ′ t^(')(t^(''))t^{\prime}\left(t^{\prime \prime}\right)t′(t′′)-axis has been marked. Since the t ′ t ′ t^(')t^{\prime}t′ and t ′ ′ t ′ ′ t^('')t^{\prime \prime}t′′ axes are tilted with the same angle relative to the t t ttt-axis,
Figure 3.3 The spacetime diagram describing the situation in Problem 1.20.
Figure 3.4 Spacetime diagrams describing the situation in Problem 1.21.
they both have the same normalization. Thus, t 2 L > t L t 2 L > t L t_(2L) > t_(L)t_{2 L}>t_{L}t2L>tL, i.e., the time measured in the rest frame of the shorter rocket is shorter.
1.21
The situation is described by the spacetime diagrams in Figure 3.4. In part (a) of the diagram, we see the situation described in S S SSS, the rest frame of the Earth. The light gray thick line (marked by μ μ mu\muμ ) corresponds to the muon worldline, the length of which is the muon eigentime to reach the Earth's surface. The black solid curve is a hyperbolic curve, which means that straight worldlines from the origin to that curve will have the same length, i.e., eigentime. From this we draw the conclusion
that the eigentime experienced by the muons will be significantly smaller than the time taken to perform the travel from the top of the atmosphere to the surface and the muons thus will not have sufficient time to decay. More quantitative statements cannot be made without further information. The t ′ t ′ t^(')t^{\prime}t′ - and x ′ x ′ x^(')x^{\prime}x′-axes of S ′ S ′ S^(')S^{\prime}S′, the muon rest frame, are also shown along with the light gray thin line, which shows the light cone. For completeness, we also show how the situation looks in S ′ S ′ S^(')S^{\prime}S′ in part (b) of the figure.
1.22
The problem is most easily solved by considering the two events when the ends of the blade reaches the y ′ = y = 0 y ′ = y = 0 y^(')=y=0y^{\prime}=y=0y′=y=0 plane. Letting the tip of the guillotine cutting the plane be the origin x = x ′ = t = t ′ = 0 x = x ′ = t = t ′ = 0 x=x^(')=t=t^(')=0x=x^{\prime}=t=t^{\prime}=0x=x′=t=t′=0, the other end of the guillotine cutting the plane will have the coordinates
(3.76) t = ℓ u = ℓ 0 u γ u , x = L , (3.76) t = ℓ u = ℓ 0 u γ u , x = L , {:(3.76)t=(ℓ)/(u)=(ℓ_(0))/(ugamma_(u))","quad x=L",":}\begin{equation*} t=\frac{\ell}{u}=\frac{\ell_{0}}{u \gamma_{u}}, \quad x=L, \tag{3.76} \end{equation*}(3.76)t=ℓu=ℓ0uγu,x=L,
where we have taken into account that the blade is length contracted in the y y yyy direction in S S SSS with a factor γ u = 1 / 1 − u 2 γ u = 1 / 1 − u 2 gamma_(u)=1//sqrt(1-u^(2))\gamma_{u}=1 / \sqrt{1-u^{2}}γu=1/1−u2. Transforming this to the S ′ S ′ S^(')S^{\prime}S′ frame, we find that
(3.77) t ′ = γ v ( t − v x ) = γ v ( ℓ 0 u γ u − v L ) (3.77) t ′ = γ v ( t − v x ) = γ v ℓ 0 u γ u − v L {:(3.77)t^(')=gamma_(v)(t-vx)=gamma_(v)((ℓ_(0))/(ugamma_(u))-vL):}\begin{equation*} t^{\prime}=\gamma_{v}(t-v x)=\gamma_{v}\left(\frac{\ell_{0}}{u \gamma_{u}}-v L\right) \tag{3.77} \end{equation*}(3.77)t′=γv(t−vx)=γv(ℓ0uγu−vL)
If the blade edge is horizontal in S ′ S ′ S^(')S^{\prime}S′, then all points on the edge will cut the y ′ = 0 y ′ = 0 y^(')=0y^{\prime}=0y′=0 plane at the same time and since t ′ = 0 t ′ = 0 t^(')=0t^{\prime}=0t′=0 for the point cutting it, we find that
(3.78) ℓ 0 u γ u − v L = 0 ⟹ v = ℓ 0 u L γ u (3.78) ℓ 0 u γ u − v L = 0 ⟹ v = ℓ 0 u L γ u {:(3.78)(ℓ_(0))/(ugamma_(u))-vL=0quad Longrightarrowquad v=(ℓ_(0))/(uLgamma_(u)):}\begin{equation*} \frac{\ell_{0}}{u \gamma_{u}}-v L=0 \quad \Longrightarrow \quad v=\frac{\ell_{0}}{u L \gamma_{u}} \tag{3.78} \end{equation*}(3.78)ℓ0uγu−vL=0⟹v=ℓ0uLγu

1.23

The spacetime diagram in Figure 3.5 shows the worldlines of the observer, the two lights, and a light signal from the lights that arrives to the observer at the same time.
The positions where the lights were when this signal was emitted will be the positions seen by the observer at t = t ∗ t = t ∗ t=t^(**)t=t^{*}t=t∗ and the seen separation L L LLL will be given by the difference of these positions.
The worldlines of the lights are given by
(3.79) x 1 ( t ) = x 1 , 0 − v t , x 2 ( t ) = x 2 , 0 − v t (3.79) x 1 ( t ) = x 1 , 0 − v t , x 2 ( t ) = x 2 , 0 − v t {:(3.79)x_(1)(t)=x_(1,0)-vt","quadx_(2)(t)=x_(2,0)-vt:}\begin{equation*} x_{1}(t)=x_{1,0}-v t, \quad x_{2}(t)=x_{2,0}-v t \tag{3.79} \end{equation*}(3.79)x1(t)=x1,0−vt,x2(t)=x2,0−vt
If the time coordinate is chosen such that the signal is emitted from x 2 ( 0 ) x 2 ( 0 ) x_(2)(0)x_{2}(0)x2(0) at t = 0 t = 0 t=0t=0t=0, the signal worldline is given by
(3.80) x s ( t ) = x 2 , 0 − t (3.80) x s ( t ) = x 2 , 0 − t {:(3.80)x_(s)(t)=x_(2,0)-t:}\begin{equation*} x_{s}(t)=x_{2,0}-t \tag{3.80} \end{equation*}(3.80)xs(t)=x2,0−t
Figure 3.5 Spacetime diagram showing the worldlines of an observer, two lights (L1 and L2), and a light signal arriving to the observer at t = t ∗ t = t ∗ t=t^(**)t=t^{*}t=t∗.
and intersects the worldlines of light 2 at x 2 , 0 x 2 , 0 x_(2,0)x_{2,0}x2,0. The intersection of the worldline of light 1 is given by
x 1 ( t 1 ) = x s ( t 1 ) ⇒ x 1 , 0 − v t 1 = x 2 , 0 − t 1 ⇒ x 2 , 0 − x 1 , 0 = ( 1 − v ) t 1 (3.81) ⇒ t 1 = x 2 , 0 − x 1 , 0 1 − v = ℓ 0 γ ( 1 − v ) = ℓ 0 1 + v 1 − v x 1 t 1 = x s t 1 ⇒ x 1 , 0 − v t 1 = x 2 , 0 − t 1 ⇒ x 2 , 0 − x 1 , 0 = ( 1 − v ) t 1 (3.81) ⇒ t 1 = x 2 , 0 − x 1 , 0 1 − v = ℓ 0 γ ( 1 − v ) = ℓ 0 1 + v 1 − v {:[x_(1)(t_(1))=x_(s)(t_(1))=>x_(1,0)-vt_(1)=x_(2,0)-t_(1)quad=>quadx_(2,0)-x_(1,0)=(1-v)t_(1)],[(3.81)=>t_(1)=(x_(2,0)-x_(1,0))/(1-v)=(ℓ_(0))/(gamma(1-v))=ℓ_(0)sqrt((1+v)/(1-v))]:}\begin{align*} x_{1}\left(t_{1}\right)=x_{s}\left(t_{1}\right) & \Rightarrow x_{1,0}-v t_{1}=x_{2,0}-t_{1} \quad \Rightarrow \quad x_{2,0}-x_{1,0}=(1-v) t_{1} \\ & \Rightarrow t_{1}=\frac{x_{2,0}-x_{1,0}}{1-v}=\frac{\ell_{0}}{\gamma(1-v)}=\ell_{0} \sqrt{\frac{1+v}{1-v}} \tag{3.81} \end{align*}x1(t1)=xs(t1)⇒x1,0−vt1=x2,0−t1⇒x2,0−x1,0=(1−v)t1(3.81)⇒t1=x2,0−x1,01−v=ℓ0γ(1−v)=ℓ01+v1−v
where we have used that x 2 , 0 − x 1 , 0 = ℓ 0 γ x 2 , 0 − x 1 , 0 = ℓ 0 γ x_(2,0)-x_(1,0)=(ℓ_(0))/(gamma)x_{2,0}-x_{1,0}=\frac{\ell_{0}}{\gamma}x2,0−x1,0=ℓ0γ by Lorentz contraction. The seen distance between the lights is therefore
(3.82) L = x s ( 0 ) − x s ( t 1 ) = x 2 , 0 − x 2 , 0 + t 1 = ℓ 0 1 + v 1 − v (3.82) L = x s ( 0 ) − x s t 1 = x 2 , 0 − x 2 , 0 + t 1 = ℓ 0 1 + v 1 − v {:(3.82)L=x_(s)(0)-x_(s)(t_(1))=x_(2,0)-x_(2,0)+t_(1)=ℓ_(0)sqrt((1+v)/(1-v)):}\begin{equation*} L=x_{s}(0)-x_{s}\left(t_{1}\right)=x_{2,0}-x_{2,0}+t_{1}=\ell_{0} \sqrt{\frac{1+v}{1-v}} \tag{3.82} \end{equation*}(3.82)L=xs(0)−xs(t1)=x2,0−x2,0+t1=ℓ01+v1−v
1.24
The Lorentz transformation representing a boost in the x x xxx-direction is given by
(3.83) c t ′ = γ ( c t − v c x ) , x ′ = γ ( x − v t ) , y ′ = y , z ′ = z (3.83) c t ′ = γ c t − v c x , x ′ = γ ( x − v t ) , y ′ = y , z ′ = z {:(3.83)ct^(')=gamma(ct-(v)/(c)x)","quadx^(')=gamma(x-vt)","quady^(')=y","quadz^(')=z:}\begin{equation*} c t^{\prime}=\gamma\left(c t-\frac{v}{c} x\right), \quad x^{\prime}=\gamma(x-v t), \quad y^{\prime}=y, \quad z^{\prime}=z \tag{3.83} \end{equation*}(3.83)ct′=γ(ct−vcx),x′=γ(x−vt),y′=y,z′=z
An object traveling at speed c c ccc in the frame S S SSS has a position given by x = c n t + x 0 x = c n t + x 0 x=cnt+x_(0)\boldsymbol{x}=c \boldsymbol{n} t+\boldsymbol{x}_{0}x=cnt+x0, where n 2 = 1 n 2 = 1 n^(2)=1\boldsymbol{n}^{2}=1n2=1. The Lorentz transform then becomes
c t ′ = γ ( c − v n 1 ) t + c t 0 ′ , x ′ = γ ( c n 1 − v ) t + x 0 ′ (3.84) y ′ = c n 2 t + y 0 ′ , z ′ = c n 3 t + z 0 ′ c t ′ = γ c − v n 1 t + c t 0 ′ , x ′ = γ c n 1 − v t + x 0 ′ (3.84) y ′ = c n 2 t + y 0 ′ , z ′ = c n 3 t + z 0 ′ {:[ct^(')=gamma(c-vn_(1))t+ct_(0)^(')","quadx^(')=gamma(cn_(1)-v)t+x_(0)^(')],[(3.84)y^(')=cn_(2)t+y_(0)^(')","quadz^(')=cn_(3)t+z_(0)^(')]:}\begin{align*} & c t^{\prime}=\gamma\left(c-v n_{1}\right) t+c t_{0}^{\prime}, \quad x^{\prime}=\gamma\left(c n_{1}-v\right) t+x_{0}^{\prime} \\ & y^{\prime}=c n_{2} t+y_{0}^{\prime}, \quad z^{\prime}=c n_{3} t+z_{0}^{\prime} \tag{3.84} \end{align*}ct′=γ(c−vn1)t+ct0′,x′=γ(cn1−v)t+x0′(3.84)y′=cn2t+y0′,z′=cn3t+z0′
The velocity in the new inertial frame S ′ S ′ S^(')S^{\prime}S′ is given by
(3.85) v ′ = d x ′ d t ′ = d x ′ / d t d t ′ / d t = ( γ ( c n 1 − v ) , c n 2 , c n 3 ) γ ( 1 − v n 1 / c ) (3.85) v ′ = d x ′ d t ′ = d x ′ / d t d t ′ / d t = γ c n 1 − v , c n 2 , c n 3 γ 1 − v n 1 / c {:(3.85)v^(')=(dx^('))/(dt^('))=(dx^(')//dt)/(dt^(')//dt)=((gamma(cn_(1)-v),cn_(2),cn_(3)))/(gamma(1-vn_(1)//c)):}\begin{equation*} \boldsymbol{v}^{\prime}=\frac{d \boldsymbol{x}^{\prime}}{d t^{\prime}}=\frac{d \boldsymbol{x}^{\prime} / d t}{d t^{\prime} / d t}=\frac{\left(\gamma\left(c n_{1}-v\right), c n_{2}, c n_{3}\right)}{\gamma\left(1-v n_{1} / c\right)} \tag{3.85} \end{equation*}(3.85)v′=dx′dt′=dx′/dtdt′/dt=(γ(cn1−v),cn2,cn3)γ(1−vn1/c)
Squaring this relation leads to
v ′ 2 = γ 2 ( c 2 n 1 2 − 2 c n 1 v + v 2 ) + c 2 ( n 2 2 + n 3 2 ) γ 2 ( 1 − 2 v n 1 / c + v 2 n 1 2 / c 2 ) = γ 2 [ c 2 n 1 2 − 2 c n 1 v + v 2 + ( c 2 − v 2 ) ( n 2 2 + n 3 2 ) ] γ 2 ( 1 − 2 v n 1 / c + v 2 n 1 2 / c 2 ) (3.86) = c 2 − 2 c n 1 v + v 2 n 1 2 1 − 2 v n 1 / c + v 2 n 1 2 / c 2 = c 2 , v ′ 2 = γ 2 c 2 n 1 2 − 2 c n 1 v + v 2 + c 2 n 2 2 + n 3 2 γ 2 1 − 2 v n 1 / c + v 2 n 1 2 / c 2 = γ 2 c 2 n 1 2 − 2 c n 1 v + v 2 + c 2 − v 2 n 2 2 + n 3 2 γ 2 1 − 2 v n 1 / c + v 2 n 1 2 / c 2 (3.86) = c 2 − 2 c n 1 v + v 2 n 1 2 1 − 2 v n 1 / c + v 2 n 1 2 / c 2 = c 2 , {:[v^('2)=(gamma^(2)(c^(2)n_(1)^(2)-2cn_(1)v+v^(2))+c^(2)(n_(2)^(2)+n_(3)^(2)))/(gamma^(2)(1-2vn_(1)//c+v^(2)n_(1)^(2)//c^(2)))],[=(gamma^(2)[c^(2)n_(1)^(2)-2cn_(1)v+v^(2)+(c^(2)-v^(2))(n_(2)^(2)+n_(3)^(2))])/(gamma^(2)(1-2vn_(1)//c+v^(2)n_(1)^(2)//c^(2)))],[(3.86)=(c^(2)-2cn_(1)v+v^(2)n_(1)^(2))/(1-2vn_(1)//c+v^(2)n_(1)^(2)//c^(2))=c^(2)","]:}\begin{align*} \boldsymbol{v}^{\prime 2} & =\frac{\gamma^{2}\left(c^{2} n_{1}^{2}-2 c n_{1} v+v^{2}\right)+c^{2}\left(n_{2}^{2}+n_{3}^{2}\right)}{\gamma^{2}\left(1-2 v n_{1} / c+v^{2} n_{1}^{2} / c^{2}\right)} \\ & =\frac{\gamma^{2}\left[c^{2} n_{1}^{2}-2 c n_{1} v+v^{2}+\left(c^{2}-v^{2}\right)\left(n_{2}^{2}+n_{3}^{2}\right)\right]}{\gamma^{2}\left(1-2 v n_{1} / c+v^{2} n_{1}^{2} / c^{2}\right)} \\ & =\frac{c^{2}-2 c n_{1} v+v^{2} n_{1}^{2}}{1-2 v n_{1} / c+v^{2} n_{1}^{2} / c^{2}}=c^{2}, \tag{3.86} \end{align*}v′2=γ2(c2n12−2cn1v+v2)+c2(n22+n32)γ2(1−2vn1/c+v2n12/c2)=γ2[c2n12−2cn1v+v2+(c2−v2)(n22+n32)]γ2(1−2vn1/c+v2n12/c2)(3.86)=c2−2cn1v+v2n121−2vn1/c+v2n12/c2=c2,
where we have used that n 1 2 + n 2 2 + n 3 2 = 1 n 1 2 + n 2 2 + n 3 2 = 1 n_(1)^(2)+n_(2)^(2)+n_(3)^(2)=1n_{1}^{2}+n_{2}^{2}+n_{3}^{2}=1n12+n22+n32=1. Thus, the object travels at speed c c ccc also in S ′ S ′ S^(')S^{\prime}S′.

1.25

Let F F FFF and R R RRR denoted the front and the rear of the train, respectively. In the rest frame of the train at the time of passing of the rear in front of the station man, we take the coordinates to be x F ′ = ( 0 , L ) x F ′ = ( 0 , L ) x_(F)^(')=(0,L)x_{F}^{\prime}=(0, L)xF′=(0,L) and x R ′ = ( 0 , 0 ) x R ′ = ( 0 , 0 ) x_(R)^(')=(0,0)x_{R}^{\prime}=(0,0)xR′=(0,0). For the stationman, these coordinates are instead x F = ( x F 0 , x F 1 ) x F = x F 0 , x F 1 x_(F)=(x_(F)^(0),x_(F)^(1))x_{F}=\left(x_{F}^{0}, x_{F}^{1}\right)xF=(xF0,xF1) and x R = ( x R 0 , x R 1 ) x R = x R 0 , x R 1 x_(R)=(x_(R)^(0),x_(R)^(1))x_{R}=\left(x_{R}^{0}, x_{R}^{1}\right)xR=(xR0,xR1), which are obtained from the first ones by means of an inverse Lorentz transformation along the x x xxx-axis, the direction of motion of the train, given by
(3.87) { x i 0 = x i ′ 0 cosh ⁡ θ + x i ′ 1 sinh ⁡ θ x i 1 = x i ′ 0 sinh ⁡ θ + x i ′ ′ cosh ⁡ θ , where i = F , R and tanh ⁡ θ = v c (3.87) x i 0 = x i ′ 0 cosh ⁡ θ + x i ′ 1 sinh ⁡ θ x i 1 = x i ′ 0 sinh ⁡ θ + x i ′ ′ cosh ⁡ θ ,  where  i = F , R  and  tanh ⁡ θ = v c {:(3.87){[x_(i)^(0)=x_(i)^('0)cosh theta+x_(i)^('1)sinh theta],[x_(i)^(1)=x_(i)^('0)sinh theta+x_(i)^('')cosh theta],quad" where "i=F,R" and "tanh theta=(v)/(c):}:}\left\{\begin{array}{l} x_{i}^{0}=x_{i}^{\prime 0} \cosh \theta+x_{i}^{\prime 1} \sinh \theta \tag{3.87}\\ x_{i}^{1}=x_{i}^{\prime 0} \sinh \theta+x_{i}^{\prime \prime} \cosh \theta \end{array}, \quad \text { where } i=F, R \text { and } \tanh \theta=\frac{v}{c}\right.(3.87){xi0=xi′0cosh⁡θ+xi′1sinh⁡θxi1=xi′0sinh⁡θ+xi′′cosh⁡θ, where i=F,R and tanh⁡θ=vc
It then holds that
(3.88) Δ x 0 = x F 0 − x R 0 = L sinh ⁡ θ = L β γ = L v c 1 1 − v 2 / c 2 (3.88) Δ x 0 = x F 0 − x R 0 = L sinh ⁡ θ = L β γ = L v c 1 1 − v 2 / c 2 {:(3.88)Deltax^(0)=x_(F)^(0)-x_(R)^(0)=L sinh theta=L beta gamma=L(v)/(c)(1)/(sqrt(1-v^(2)//c^(2))):}\begin{equation*} \Delta x^{0}=x_{F}^{0}-x_{R}^{0}=L \sinh \theta=L \beta \gamma=L \frac{v}{c} \frac{1}{\sqrt{1-v^{2} / c^{2}}} \tag{3.88} \end{equation*}(3.88)Δx0=xF0−xR0=Lsinh⁡θ=Lβγ=Lvc11−v2/c2
The time difference, in the rest frame of the stationman, is therefore
(3.89) Δ t = Δ x 0 c = v L c 2 1 − v 2 / c 2 (3.89) Δ t = Δ x 0 c = v L c 2 1 − v 2 / c 2 {:(3.89)Delta t=(Deltax^(0))/(c)=(vL)/(c^(2)sqrt(1-v^(2)//c^(2))):}\begin{equation*} \Delta t=\frac{\Delta x^{0}}{c}=\frac{v L}{c^{2} \sqrt{1-v^{2} / c^{2}}} \tag{3.89} \end{equation*}(3.89)Δt=Δx0c=vLc21−v2/c2

1.26

Let A A AAA and B B BBB be the front and rear end of the train, respectively. Since the velocity of light is c c ccc for all observers, the times are given by t 1 = − L x / c t 1 = − L x / c t_(1)=-Lx//ct_{1}=-L x / ct1=−Lx/c and t 2 = − L ( 1 − x ) / c t 2 = − L ( 1 − x ) / c t_(2)=-L(1-x)//ct_{2}=-L(1-x) / ct2=−L(1−x)/c. These times are related to the times t 1 ′ t 1 ′ t_(1)^(')t_{1}^{\prime}t1′ and t 2 ′ t 2 ′ t_(2)^(')t_{2}^{\prime}t2′ determined by the observer on the ground by a Lorentz transformation
(3.90) c t 2 ′ = c t 2 cosh ⁡ θ + [ − L ( 1 − x ) ] sinh ⁡ θ (3.91) c t 1 ′ = c t 1 cosh ⁡ θ + x L sinh ⁡ θ (3.90) c t 2 ′ = c t 2 cosh ⁡ θ + [ − L ( 1 − x ) ] sinh ⁡ θ (3.91) c t 1 ′ = c t 1 cosh ⁡ θ + x L sinh ⁡ θ {:[(3.90)ct_(2)^(')=ct_(2)cosh theta+[-L(1-x)]sinh theta],[(3.91)ct_(1)^(')=ct_(1)cosh theta+xL sinh theta]:}\begin{align*} & c t_{2}^{\prime}=c t_{2} \cosh \theta+[-L(1-x)] \sinh \theta \tag{3.90}\\ & c t_{1}^{\prime}=c t_{1} \cosh \theta+x L \sinh \theta \tag{3.91} \end{align*}(3.90)ct2′=ct2cosh⁡θ+[−L(1−x)]sinh⁡θ(3.91)ct1′=ct1cosh⁡θ+xLsinh⁡θ
where we have put the origin at O = O ′ O = O ′ O=O^(')O=O^{\prime}O=O′. This gives c ( t 2 ′ − t 1 ′ ) = c ( t 2 − t 1 ) cosh ⁡ θ − c t 2 ′ − t 1 ′ = c t 2 − t 1 cosh ⁡ θ − c(t_(2)^(')-t_(1)^('))=c(t_(2)-t_(1))cosh theta-c\left(t_{2}^{\prime}-t_{1}^{\prime}\right)=c\left(t_{2}-t_{1}\right) \cosh \theta-c(t2′−t1′)=c(t2−t1)cosh⁡θ− L sinh ⁡ θ L sinh ⁡ θ L sinh thetaL \sinh \thetaLsinh⁡θ. If t 2 ′ − t 1 ′ = 0 t 2 ′ − t 1 ′ = 0 t_(2)^(')-t_(1)^(')=0t_{2}^{\prime}-t_{1}^{\prime}=0t2′−t1′=0, we obtain v = c tanh ⁡ θ = ( 2 x − 1 ) c v = c tanh ⁡ θ = ( 2 x − 1 ) c v=c tanh theta=(2x-1)cv=c \tanh \theta=(2 x-1) cv=ctanh⁡θ=(2x−1)c, and v = 0 v = 0 v=0v=0v=0 therefore
implies x = 1 / 2 x = 1 / 2 x=1//2x=1 / 2x=1/2. This means that if x < 1 / 2 x < 1 / 2 x < 1//2x<1 / 2x<1/2 the train has to move in opposite direction to when x > 1 / 2 x > 1 / 2 x > 1//2x>1 / 2x>1/2, for the situation to occur, i.e., A A AAA and B B BBB change roles of being rear and front, respectively.
One can also calculate the invariant interval s 2 = c 2 ( t 2 − t 1 ) 2 − L 2 = c 2 ( t 2 ′ − s 2 = c 2 t 2 − t 1 2 − L 2 = c 2 t 2 ′ − s^(2)=c^(2)(t_(2)-t_(1))^(2)-L^(2)=c^(2)(t_(2)^(')-:}s^{2}=c^{2}\left(t_{2}-t_{1}\right)^{2}-L^{2}=c^{2}\left(t_{2}^{\prime}-\right.s2=c2(t2−t1)2−L2=c2(t2′− t 1 ′ ) 2 − L ′ 2 t 1 ′ 2 − L ′ 2 t_(1)^('))^(2)-L^('2)\left.t_{1}^{\prime}\right)^{2}-L^{\prime 2}t1′)2−L′2 in the two frames and use the length contraction formula to obtain the velocity. In this treatment, the sign of the velocity must be discussed separately.
1.27
a) Suppose the particle moves through the origin of K K KKK. Then, the event A = A = A=A=A= ( c t , 0 , 0 , − u t c t , 0 , 0 , − u t ct,0,0,-utc t, 0,0,-u tct,0,0,−ut ) belongs to the worldline of the particle. Transforming A A AAA to K ′ K ′ K^(')K^{\prime}K′ using the standard Lorentz transformation, we find that A ′ = ( ct γ ( v ) , − v t γ ( v ) , 0 , − u t ) A ′ = ( ct γ ( v ) , − v t γ ( v ) , 0 , − u t ) A^(')=(ct gamma(v),-vt gamma(v),0,-ut)A^{\prime}=(\operatorname{ct\gamma }(v),-v t \gamma(v), 0,-u t)A′=(ctγ(v),−vtγ(v),0,−ut). The angle is therefore given by
(3.92) tan ⁡ θ = γ ( v ) v u ⇒ θ = arctan ⁡ ( γ ( v ) v u ) (3.92) tan ⁡ θ = γ ( v ) v u ⇒ θ = arctan ⁡ γ ( v ) v u {:(3.92)tan theta=gamma(v)(v)/(u)=>theta=arctan(gamma(v)(v)/(u)):}\begin{equation*} \tan \theta=\gamma(v) \frac{v}{u} \Rightarrow \theta=\arctan \left(\gamma(v) \frac{v}{u}\right) \tag{3.92} \end{equation*}(3.92)tan⁡θ=γ(v)vu⇒θ=arctan⁡(γ(v)vu)
b) The stars will all seem to gather in front of the spaceship.

1.28

The Lorentz transformation of the 4 -velocity to the new system gives for the nontrivial components
(3.93) V ′ 0 = c γ ( v ′ ′ ) = c γ ( v ′ ) cosh ⁡ θ + γ ( v ′ ) v ′ sinh ⁡ θ (3.94) V ′ 1 = v ′ ′ γ ( v ′ ′ ) = c γ ( v ′ ) sinh ⁡ θ + γ ( v ′ ) v ′ cosh ⁡ θ (3.93) V ′ 0 = c γ v ′ ′ = c γ v ′ cosh ⁡ θ + γ v ′ v ′ sinh ⁡ θ (3.94) V ′ 1 = v ′ ′ γ v ′ ′ = c γ v ′ sinh ⁡ θ + γ v ′ v ′ cosh ⁡ θ {:[(3.93)V^('0)=c gamma(v^(''))=c gamma(v^('))cosh theta+gamma(v^('))v^(')sinh theta],[(3.94)V^('1)=v^('')gamma(v^(''))=c gamma(v^('))sinh theta+gamma(v^('))v^(')cosh theta]:}\begin{align*} V^{\prime 0}=c \gamma\left(v^{\prime \prime}\right) & =c \gamma\left(v^{\prime}\right) \cosh \theta+\gamma\left(v^{\prime}\right) v^{\prime} \sinh \theta \tag{3.93}\\ V^{\prime 1}=v^{\prime \prime} \gamma\left(v^{\prime \prime}\right) & =c \gamma\left(v^{\prime}\right) \sinh \theta+\gamma\left(v^{\prime}\right) v^{\prime} \cosh \theta \tag{3.94} \end{align*}(3.93)V′0=cγ(v′′)=cγ(v′)cosh⁡θ+γ(v′)v′sinh⁡θ(3.94)V′1=v′′γ(v′′)=cγ(v′)sinh⁡θ+γ(v′)v′cosh⁡θ
where v = c tanh ⁡ θ v = c tanh ⁡ θ v=c tanh thetav=c \tanh \thetav=ctanh⁡θ. Calculating v ′ ′ = V ′ 1 / v ′ 0 v ′ ′ = V ′ 1 / v ′ 0 v^('')=V^('1)//v^('0)v^{\prime \prime}=V^{\prime 1} / v^{\prime 0}v′′=V′1/v′0 yields
(3.95) v ′ ′ = c sinh ⁡ θ + v ′ c ⋅ cosh ⁡ θ cosh ⁡ θ + v ′ c sinh ⁡ θ = v + v ′ 1 + v v ′ c 2 (3.95) v ′ ′ = c sinh ⁡ θ + v ′ c ⋅ cosh ⁡ θ cosh ⁡ θ + v ′ c sinh ⁡ θ = v + v ′ 1 + v v ′ c 2 {:(3.95)v^('')=c(sinh theta+(v^('))/(c)*cosh theta)/(cosh theta+(v^('))/(c)sinh theta)=(v+v^('))/(1+(vv^('))/(c^(2))):}\begin{equation*} v^{\prime \prime}=c \frac{\sinh \theta+\frac{v^{\prime}}{c} \cdot \cosh \theta}{\cosh \theta+\frac{v^{\prime}}{c} \sinh \theta}=\frac{v+v^{\prime}}{1+\frac{v v^{\prime}}{c^{2}}} \tag{3.95} \end{equation*}(3.95)v′′=csinh⁡θ+v′c⋅cosh⁡θcosh⁡θ+v′csinh⁡θ=v+v′1+vv′c2
which is the desired formula for relativistic addition of velocities.

1.29

The standard configuration Lorentz transformation (in one temporal and two spatial dimensions) is given by x ′ = Λ x x ′ = Λ x x^(')=Lambda xx^{\prime}=\Lambda xx′=Λx, where
(3.96) x = ( x 0 x 1 x 2 ) and Λ = ( γ ( v ) − v c γ ( v ) 0 − v c γ ( v ) γ ( v ) 0 0 0 1 ) . (3.96) x = x 0 x 1 x 2  and  Λ = γ ( v ) − v c γ ( v ) 0 − v c γ ( v ) γ ( v ) 0 0 0 1 . {:(3.96)x=([x^(0)],[x^(1)],[x^(2)])quad" and "quad Lambda=([gamma(v),-(v)/(c)gamma(v),0],[-(v)/(c)gamma(v),gamma(v),0],[0,0,1]).:}x=\left(\begin{array}{l} x^{0} \tag{3.96}\\ x^{1} \\ x^{2} \end{array}\right) \quad \text { and } \quad \Lambda=\left(\begin{array}{ccc} \gamma(v) & -\frac{v}{c} \gamma(v) & 0 \\ -\frac{v}{c} \gamma(v) & \gamma(v) & 0 \\ 0 & 0 & 1 \end{array}\right) .(3.96)x=(x0x1x2) and Λ=(γ(v)−vcγ(v)0−vcγ(v)γ(v)0001).
Here γ ( v ) = 1 1 − ( v c ) 2 γ ( v ) = 1 1 − v c 2 gamma(v)=(1)/(sqrt(1-((v)/(c))^(2)))\gamma(v)=\frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^{2}}}γ(v)=11−(vc)2. This means that
(3.97) x ′ 0 = γ ( v ) ( x 0 − v c x 1 ) , (3.97) x ′ 0 = γ ( v ) x 0 − v c x 1 , {:(3.97)x^('0)=gamma(v)(x^(0)-(v)/(c)x^(1))",":}\begin{equation*} x^{\prime 0}=\gamma(v)\left(x^{0}-\frac{v}{c} x^{1}\right), \tag{3.97} \end{equation*}(3.97)x′0=γ(v)(x0−vcx1),
(3.98) x ′ 1 = γ ( v ) ( x 1 − v c x 0 ) , (3.99) x ′ 2 = x 2 . (3.98) x ′ 1 = γ ( v ) x 1 − v c x 0 , (3.99) x ′ 2 = x 2 . {:[(3.98)x^('1)=gamma(v)(x^(1)-(v)/(c)x^(0))","],[(3.99)x^('2)=x^(2).]:}\begin{align*} & x^{\prime 1}=\gamma(v)\left(x^{1}-\frac{v}{c} x^{0}\right), \tag{3.98}\\ & x^{\prime 2}=x^{2} . \tag{3.99} \end{align*}(3.98)x′1=γ(v)(x1−vcx0),(3.99)x′2=x2.
The observer in K ′ K ′ K^(')K^{\prime}K′ measures the triangle at time x ′ 0 = 0 x ′ 0 = 0 x^('0)=0x^{\prime 0}=0x′0=0. Using Eq. (3.97) together with x ′ 0 = 0 x ′ 0 = 0 x^('0)=0x^{\prime 0}=0x′0=0, implies that x 0 = v c x 1 x 0 = v c x 1 x^(0)=(v)/(c)x^(1)x^{0}=\frac{v}{c} x^{1}x0=vcx1. Inserting x 0 = v c x 1 x 0 = v c x 1 x^(0)=(v)/(c)x^(1)x^{0}=\frac{v}{c} x^{1}x0=vcx1 into Eq. (3.98), yields
(3.100) x ′ 1 = γ ( v ) ( x 1 − v c ⋅ v c x 1 ) = γ ( v ) ( 1 − v 2 c 2 ) x 1 = γ ( v ) 1 γ ( v ) 2 x 1 = 1 γ ( v ) x 1 (3.100) x ′ 1 = γ ( v ) x 1 − v c ⋅ v c x 1 = γ ( v ) 1 − v 2 c 2 x 1 = γ ( v ) 1 γ ( v ) 2 x 1 = 1 γ ( v ) x 1 {:(3.100)x^('1)=gamma(v)(x^(1)-(v)/(c)*(v)/(c)x^(1))=gamma(v)(1-(v^(2))/(c^(2)))x^(1)=gamma(v)(1)/(gamma(v)^(2))x^(1)=(1)/(gamma(v))x^(1):}\begin{equation*} x^{\prime 1}=\gamma(v)\left(x^{1}-\frac{v}{c} \cdot \frac{v}{c} x^{1}\right)=\gamma(v)\left(1-\frac{v^{2}}{c^{2}}\right) x^{1}=\gamma(v) \frac{1}{\gamma(v)^{2}} x^{1}=\frac{1}{\gamma(v)} x^{1} \tag{3.100} \end{equation*}(3.100)x′1=γ(v)(x1−vc⋅vcx1)=γ(v)(1−v2c2)x1=γ(v)1γ(v)2x1=1γ(v)x1
i.e., x ′ 1 = 1 γ ( v ) x 1 x ′ 1 = 1 γ ( v ) x 1 x^('1)=(1)/(gamma(v))x^(1)x^{\prime 1}=\frac{1}{\gamma(v)} x^{1}x′1=1γ(v)x1, which is the Lorentz length contraction formula.
In K K KKK :
All three sides of the triangle have length ℓ ℓ ℓ\ellℓ and all three angles in the triangle are 60 ∘ 60 ∘ 60^(@)60^{\circ}60∘ (i.e., π 3 π 3 (pi)/(3)\frac{\pi}{3}π3 ). One of the sides in the triangle (the base b = ℓ b = ℓ b=ℓb=\ellb=ℓ ) is parallel to the x 1 x 1 x^(1)x^{1}x1-axis. Using Pythagoras' theorem, ℓ 2 = ( b 2 ) 2 + h 2 ℓ 2 = b 2 2 + h 2 ℓ^(2)=((b)/(2))^(2)+h^(2)\ell^{2}=\left(\frac{b}{2}\right)^{2}+h^{2}ℓ2=(b2)2+h2, one finds the length of the altitude (the height) of the triangle to be h = 3 2 ⋅ ℓ h = 3 2 ⋅ ℓ h=(sqrt3)/(2)*ℓh=\frac{\sqrt{3}}{2} \cdot \ellh=32⋅ℓ.
In K ′ K ′ K^(')K^{\prime}K′ :
The length of base of the triangle is: b ′ = 1 γ ( v ) ⋅ b = 1 γ ( v ) ℓ b ′ = 1 γ ( v ) ⋅ b = 1 γ ( v ) ℓ b^(')=(1)/(gamma(v))*b=(1)/(gamma(v))ℓb^{\prime}=\frac{1}{\gamma(v)} \cdot b=\frac{1}{\gamma(v)} \ellb′=1γ(v)⋅b=1γ(v)ℓ. The length of the altitude (the height) of the triangle is: h ′ = h = 3 2 ℓ h ′ = h = 3 2 ℓ h^(')=h=(sqrt3)/(2)ℓh^{\prime}=h=\frac{\sqrt{3}}{2} \ellh′=h=32ℓ [using Eq. (3.99)]. Assume that the length of the two other sides of the triangle is ℓ ′ ℓ ′ ℓ^(')\ell^{\prime}ℓ′. Again, using Pythagoras' theorem, ℓ ′ 2 = ( b ′ 2 ) 2 + h ′ 2 ℓ ′ 2 = b ′ 2 2 + h ′ 2 ℓ^('2)=((b^('))/(2))^(2)+h^('2)\ell^{\prime 2}=\left(\frac{b^{\prime}}{2}\right)^{2}+h^{\prime 2}ℓ′2=(b′2)2+h′2, one finds the length of the two other sides as
(3.101) ℓ ′ = ( b ′ 2 ) 2 + h ′ 2 = [ ℓ 2 γ ( v ) ] 2 + ( 3 2 ℓ ) 2 = ℓ 2 3 + 1 γ ( v ) 2 (3.101) ℓ ′ = b ′ 2 2 + h ′ 2 = ℓ 2 γ ( v ) 2 + 3 2 ℓ 2 = ℓ 2 3 + 1 γ ( v ) 2 {:(3.101)ℓ^(')=sqrt(((b^('))/(2))^(2)+h^('2))=sqrt([(ℓ)/(2gamma(v))]^(2)+((sqrt3)/(2)ℓ)^(2))=(ℓ)/(2)sqrt(3+(1)/(gamma(v)^(2))):}\begin{equation*} \ell^{\prime}=\sqrt{\left(\frac{b^{\prime}}{2}\right)^{2}+h^{\prime 2}}=\sqrt{\left[\frac{\ell}{2 \gamma(v)}\right]^{2}+\left(\frac{\sqrt{3}}{2} \ell\right)^{2}}=\frac{\ell}{2} \sqrt{3+\frac{1}{\gamma(v)^{2}}} \tag{3.101} \end{equation*}(3.101)ℓ′=(b′2)2+h′2=[ℓ2γ(v)]2+(32ℓ)2=ℓ23+1γ(v)2
The base angle α α alpha\alphaα can be obtained from the relation ℓ ′ ⋅ cos ⁡ α = b ′ 2 ℓ ′ ⋅ cos ⁡ α = b ′ 2 ℓ^(')*cos alpha=(b^('))/(2)\ell^{\prime} \cdot \cos \alpha=\frac{b^{\prime}}{2}ℓ′⋅cos⁡α=b′2 and the apex angle β β beta\betaβ from the relation ℓ ′ ⋅ sin ⁡ β 2 = b ′ 2 ℓ ′ ⋅ sin ⁡ β 2 = b ′ 2 ℓ^(')*sin((beta)/(2))=(b^('))/(2)\ell^{\prime} \cdot \sin \frac{\beta}{2}=\frac{b^{\prime}}{2}ℓ′⋅sin⁡β2=b′2. The results are: α = arccos ⁡ 1 1 + 3 γ ( v ) 2 α = arccos ⁡ 1 1 + 3 γ ( v ) 2 alpha=arccos((1)/(sqrt(1+3gamma(v)^(2))))\alpha=\arccos \frac{1}{\sqrt{1+3 \gamma(v)^{2}}}α=arccos⁡11+3γ(v)2 and β = 2 arcsin ⁡ 1 1 + 3 γ ( v ) 2 β = 2 arcsin ⁡ 1 1 + 3 γ ( v ) 2 beta=2arcsin((1)/(sqrt(1+3gamma(v)^(2))))\beta=2 \arcsin \frac{1}{\sqrt{1+3 \gamma(v)^{2}}}β=2arcsin⁡11+3γ(v)2.

1.30

For two events, E 1 E 1 E_(1)E_{1}E1 and E 2 E 2 E_(2)E_{2}E2 that occur at the left and right endpoints of the rod, respectively, we have
(3.102) Δ x ′ 2 = u Δ t ′ (3.102) Δ x ′ 2 = u Δ t ′ {:(3.102)Deltax^('2)=u Deltat^('):}\begin{equation*} \Delta x^{\prime 2}=u \Delta t^{\prime} \tag{3.102} \end{equation*}(3.102)Δx′2=uΔt′
By using the standard configuration Lorentz transformation, this can reformulated as
(3.103) Δ x 2 = u γ ( v ) ( Δ t − v c 2 ⋅ Δ x 1 ) (3.103) Δ x 2 = u γ ( v ) Δ t − v c 2 ⋅ Δ x 1 {:(3.103)Deltax^(2)=u gamma(v)(Delta t-(v)/(c^(2))*Deltax^(1)):}\begin{equation*} \Delta x^{2}=u \gamma(v)\left(\Delta t-\frac{v}{c^{2}} \cdot \Delta x^{1}\right) \tag{3.103} \end{equation*}(3.103)Δx2=uγ(v)(Δt−vc2⋅Δx1)
When Δ t = 0 Δ t = 0 Delta t=0\Delta t=0Δt=0, we obtain
(3.104) tan ⁡ ϕ = Δ x 2 Δ x 1 = − u v c 2 γ ( v ) = − u v / c 2 1 − v 2 / c 2 (3.104) tan ⁡ ϕ = Δ x 2 Δ x 1 = − u v c 2 γ ( v ) = − u v / c 2 1 − v 2 / c 2 {:(3.104)tan phi=(Deltax^(2))/(Deltax^(1))=-(uv)/(c^(2))gamma(v)=-(uv//c^(2))/(sqrt(1-v^(2)//c^(2))):}\begin{equation*} \tan \phi=\frac{\Delta x^{2}}{\Delta x^{1}}=-\frac{u v}{c^{2}} \gamma(v)=-\frac{u v / c^{2}}{\sqrt{1-v^{2} / c^{2}}} \tag{3.104} \end{equation*}(3.104)tan⁡ϕ=Δx2Δx1=−uvc2γ(v)=−uv/c21−v2/c2

1.31

Let the axis of the cylinder coincide with the x x xxx-axis in K K KKK. The straight line on the cylinder surface is described by the equation
(3.105) φ = ω t (3.105) φ = ω t {:(3.105)varphi=omega t:}\begin{equation*} \varphi=\omega t \tag{3.105} \end{equation*}(3.105)φ=ωt
where φ φ varphi\varphiφ is the angle of rotation around the x x xxx-axis. We now transform the equation φ = ω t φ = ω t varphi=omega t\varphi=\omega tφ=ωt to K ′ K ′ K^(')K^{\prime}K′, that moves with velocity v v vvv in the x x xxx-direction. We find that
(3.106) φ = φ ′ and t = γ ( v ) ( t ′ + v x ′ c 2 ) (3.106) φ = φ ′  and  t = γ ( v ) t ′ + v x ′ c 2 {:(3.106)varphi=varphi^(')quad" and "quad t=gamma(v)(t^(')+(vx^('))/(c^(2))):}\begin{equation*} \varphi=\varphi^{\prime} \quad \text { and } \quad t=\gamma(v)\left(t^{\prime}+\frac{v x^{\prime}}{c^{2}}\right) \tag{3.106} \end{equation*}(3.106)φ=φ′ and t=γ(v)(t′+vx′c2)
which means that the equation of motion of the straight line relative to K ′ K ′ K^(')K^{\prime}K′ is described by the equation
(3.107) φ ′ = ω γ ( v ) ( t ′ + v x ′ c 2 ) (3.107) φ ′ = ω γ ( v ) t ′ + v x ′ c 2 {:(3.107)varphi^(')=omega gamma(v)(t^(')+(vx^('))/(c^(2))):}\begin{equation*} \varphi^{\prime}=\omega \gamma(v)\left(t^{\prime}+\frac{v x^{\prime}}{c^{2}}\right) \tag{3.107} \end{equation*}(3.107)φ′=ωγ(v)(t′+vx′c2)
The twist per unit length for a fixed t ′ t ′ t^(')t^{\prime}t′ is therefore given by
(3.108) ∂ φ ′ ∂ x ′ = ω γ ( v ) v c 2 (3.108) ∂ φ ′ ∂ x ′ = ω γ ( v ) v c 2 {:(3.108)(delvarphi^('))/(delx^('))=omega gamma(v)(v)/(c^(2)):}\begin{equation*} \frac{\partial \varphi^{\prime}}{\partial x^{\prime}}=\omega \gamma(v) \frac{v}{c^{2}} \tag{3.108} \end{equation*}(3.108)∂φ′∂x′=ωγ(v)vc2
so that to the observer in K ′ K ′ K^(')K^{\prime}K′, the straight line appears as a twisted line around the cylinder.

1.32

Let E 1 E 1 E_(1)E_{1}E1 and E 2 E 2 E_(2)E_{2}E2 be the events of turning on two compartment lights. If K K KKK and K ′ K ′ K^(')K^{\prime}K′ are the rest frames of the station and the train, respectively, then
(3.109) Δ t ′ = γ ( v ) ( Δ t − v c 2 Δ x ) = 0 (3.109) Δ t ′ = γ ( v ) Δ t − v c 2 Δ x = 0 {:(3.109)Deltat^(')=gamma(v)(Delta t-(v)/(c^(2))Delta x)=0:}\begin{equation*} \Delta t^{\prime}=\gamma(v)\left(\Delta t-\frac{v}{c^{2}} \Delta x\right)=0 \tag{3.109} \end{equation*}(3.109)Δt′=γ(v)(Δt−vc2Δx)=0
However, it holds that Δ x = u Δ t Δ x = u Δ t Delta x=u Delta t\Delta x=u \Delta tΔx=uΔt, which gives u = c 2 / v u = c 2 / v u=c^(2)//vu=c^{2} / vu=c2/v.

1.33

Let K K KKK and K ′ K ′ K^(')K^{\prime}K′ be the rest frames of the star and spaceship, respectively. Furthermore, let the planet have its orbit in the x y x y xyx yxy-plane in the coordinate system K K KKK,
i.e., z = 0 z = 0 z=0z=0z=0 for the planet. The spacetime trajectory of the planet in K K KKK is then x = ( x 0 , x ) = ( c t , x ) x = x 0 , x = ( c t , x ) x=(x^(0),x)=(ct,x)x=\left(x^{0}, \boldsymbol{x}\right)=(c t, \boldsymbol{x})x=(x0,x)=(ct,x), where x = ( R cos ⁡ ω t , R sin ⁡ ω t , 0 ) x = ( R cos ⁡ ω t , R sin ⁡ ω t , 0 ) x=(R cos omega t,R sin omega t,0)\boldsymbol{x}=(R \cos \omega t, R \sin \omega t, 0)x=(Rcos⁡ωt,Rsin⁡ωt,0). The orbit of the planet in K ′ K ′ K^(')K^{\prime}K′ is now given by the Lorentz transformation:
(3.110) { t ′ = γ ( v ) ( t − v c 2 z ) x ′ = x y ′ = y z ′ = γ ( v ) ( z − v t ) (3.110) t ′ = γ ( v ) t − v c 2 z x ′ = x y ′ = y z ′ = γ ( v ) ( z − v t ) {:(3.110){[t^(')=gamma(v)(t-(v)/(c^(2))z)],[x^(')=x],[y^(')=y],[z^(')=gamma(v)(z-vt)]:}:}\left\{\begin{array}{l} t^{\prime}=\gamma(v)\left(t-\frac{v}{c^{2}} z\right) \tag{3.110}\\ x^{\prime}=x \\ y^{\prime}=y \\ z^{\prime}=\gamma(v)(z-v t) \end{array}\right.(3.110){t′=γ(v)(t−vc2z)x′=xy′=yz′=γ(v)(z−vt)
where γ ( v ) = 1 1 − v 2 / c 2 γ ( v ) = 1 1 − v 2 / c 2 gamma(v)=(1)/(sqrt(1-v^(2)//c^(2)))\gamma(v)=\frac{1}{\sqrt{1-v^{2} / c^{2}}}γ(v)=11−v2/c2. The spaceship is moving along the positive z z zzz-axis with velocity v v vvv. Therefore, we have
(3.111) t ′ = γ ( v ) t (3.112) x ′ = R cos ⁡ ω t = R cos ⁡ ω t ′ γ ( v ) = R cos ⁡ ω ′ t ′ (3.113) y ′ = R sin ⁡ ω t = R sin ⁡ ω t ′ γ ( v ) = R sin ⁡ ω ′ t ′ (3.114) z ′ = − γ ( v ) v t = − v t ′ (3.111) t ′ = γ ( v ) t (3.112) x ′ = R cos ⁡ ω t = R cos ⁡ ω t ′ γ ( v ) = R cos ⁡ ω ′ t ′ (3.113) y ′ = R sin ⁡ ω t = R sin ⁡ ω t ′ γ ( v ) = R sin ⁡ ω ′ t ′ (3.114) z ′ = − γ ( v ) v t = − v t ′ {:[(3.111)t^(')=gamma(v)t],[(3.112)x^(')=R cos omega t=R cos omega(t^('))/(gamma(v))=R cos omega^(')t^(')],[(3.113)y^(')=R sin omega t=R sin omega(t^('))/(gamma(v))=R sin omega^(')t^(')],[(3.114)z^(')=-gamma(v)vt=-vt^(')]:}\begin{align*} & t^{\prime}=\gamma(v) t \tag{3.111}\\ & x^{\prime}=R \cos \omega t=R \cos \omega \frac{t^{\prime}}{\gamma(v)}=R \cos \omega^{\prime} t^{\prime} \tag{3.112}\\ & y^{\prime}=R \sin \omega t=R \sin \omega \frac{t^{\prime}}{\gamma(v)}=R \sin \omega^{\prime} t^{\prime} \tag{3.113}\\ & z^{\prime}=-\gamma(v) v t=-v t^{\prime} \tag{3.114} \end{align*}(3.111)t′=γ(v)t(3.112)x′=Rcos⁡ωt=Rcos⁡ωt′γ(v)=Rcos⁡ω′t′(3.113)y′=Rsin⁡ωt=Rsin⁡ωt′γ(v)=Rsin⁡ω′t′(3.114)z′=−γ(v)vt=−vt′
where ω ′ = ω / γ ( v ) ω ′ = ω / γ ( v ) omega^(')=omega//gamma(v)\omega^{\prime}=\omega / \gamma(v)ω′=ω/γ(v). Thus, the spacetime trajectory of the planet in K ′ K ′ K^(')K^{\prime}K′ is given by x ′ = ( x ′ 0 , x ′ ) = ( c t ′ , x ′ ) x ′ = x ′ 0 , x ′ = c t ′ , x ′ x^(')=(x^('0),x^('))=(ct^('),x^('))x^{\prime}=\left(x^{\prime 0}, \boldsymbol{x}^{\prime}\right)=\left(c t^{\prime}, \boldsymbol{x}^{\prime}\right)x′=(x′0,x′)=(ct′,x′), where
(3.115) x ′ ( t ′ ) = ( R cos ⁡ ω t ′ γ ( v ) , R sin ⁡ ω t ′ γ ( v ) , − v t ′ ) (3.115) x ′ t ′ = R cos ⁡ ω t ′ γ ( v ) , R sin ⁡ ω t ′ γ ( v ) , − v t ′ {:(3.115)x^(')(t^('))=(R cos((omegat^('))/(gamma(v))),R sin((omegat^('))/(gamma(v))),-vt^(')):}\begin{equation*} \boldsymbol{x}^{\prime}\left(t^{\prime}\right)=\left(R \cos \frac{\omega t^{\prime}}{\gamma(v)}, R \sin \frac{\omega t^{\prime}}{\gamma(v)},-v t^{\prime}\right) \tag{3.115} \end{equation*}(3.115)x′(t′)=(Rcos⁡ωt′γ(v),Rsin⁡ωt′γ(v),−vt′)

1.34

Let x μ , x ′ μ x μ , x ′ μ x^(mu),x^('mu)x^{\mu}, x^{\prime \mu}xμ,x′μ, and x ′ ′ μ x ′ ′ μ x^(''mu)x^{\prime \prime \mu}x′′μ be the rest coordinates of the observers A , B A , B A,BA, BA,B, and C C CCC, respectively.
The Lorentz transformation between A A AAA and B B BBB ( B B BBB is moving with velocity v v vvv along the positive x 1 x 1 x^(1)x^{1}x1-axis in K K KKK ) is given by
(3.116) { x ′ 0 = x 0 cosh ⁡ θ − x 1 sinh ⁡ θ x ′ 1 = − x 0 sinh ⁡ θ + x 1 cosh ⁡ θ x ′ 2 = x 2 x ′ 3 = x 3 (3.116) x ′ 0 = x 0 cosh ⁡ θ − x 1 sinh ⁡ θ x ′ 1 = − x 0 sinh ⁡ θ + x 1 cosh ⁡ θ x ′ 2 = x 2 x ′ 3 = x 3 {:(3.116){[x^('0)=x^(0)cosh theta-x^(1)sinh theta],[x^('1)=-x^(0)sinh theta+x^(1)cosh theta],[x^('2)=x^(2)],[x^('3)=x^(3)]:}:}\left\{\begin{array}{l} x^{\prime 0}=x^{0} \cosh \theta-x^{1} \sinh \theta \tag{3.116}\\ x^{\prime 1}=-x^{0} \sinh \theta+x^{1} \cosh \theta \\ x^{\prime 2}=x^{2} \\ x^{\prime 3}=x^{3} \end{array}\right.(3.116){x′0=x0cosh⁡θ−x1sinh⁡θx′1=−x0sinh⁡θ+x1cosh⁡θx′2=x2x′3=x3
where tanh ⁡ θ = v c tanh ⁡ θ = v c tanh theta=(v)/(c)\tanh \theta=\frac{v}{c}tanh⁡θ=vc, and the Lorentz transformation between B B BBB and C C CCC ( C C CCC is moving with velocity v ′ v ′ v^(')v^{\prime}v′ along the positive x ′ 2 x ′ 2 x^('2)x^{\prime 2}x′2-axis in K ′ K ′ K^(')K^{\prime}K′ ) is given by
(3.117) { x ′ ′ 0 = x ′ 0 cosh ⁡ θ ′ − x ′ 2 sinh ⁡ θ ′ x ′ ′ 1 = x ′ 1 x ′ ′ 2 = − x ′ 0 sinh ⁡ θ ′ + x ′ 2 cosh ⁡ θ ′ x ′ ′ 3 = x ′ 3 (3.117) x ′ ′ 0 = x ′ 0 cosh ⁡ θ ′ − x ′ 2 sinh ⁡ θ ′ x ′ ′ 1 = x ′ 1 x ′ ′ 2 = − x ′ 0 sinh ⁡ θ ′ + x ′ 2 cosh ⁡ θ ′ x ′ ′ 3 = x ′ 3 {:(3.117){[x^(''0)=x^('0)cosh theta^(')-x^('2)sinh theta^(')],[x^(''1)=x^('1)],[x^(''2)=-x^('0)sinh theta^(')+x^('2)cosh theta^(')],[x^(''3)=x^('3)]:}:}\left\{\begin{array}{l} x^{\prime \prime 0}=x^{\prime 0} \cosh \theta^{\prime}-x^{\prime 2} \sinh \theta^{\prime} \tag{3.117}\\ x^{\prime \prime 1}=x^{\prime 1} \\ x^{\prime \prime 2}=-x^{\prime 0} \sinh \theta^{\prime}+x^{\prime 2} \cosh \theta^{\prime} \\ x^{\prime \prime 3}=x^{\prime 3} \end{array}\right.(3.117){x′′0=x′0cosh⁡θ′−x′2sinh⁡θ′x′′1=x′1x′′2=−x′0sinh⁡θ′+x′2cosh⁡θ′x′′3=x′3
where tanh ⁡ θ ′ = v ′ c tanh ⁡ θ ′ = v ′ c tanh theta^(')=(v^('))/(c)\tanh \theta^{\prime}=\frac{v^{\prime}}{c}tanh⁡θ′=v′c.
Inserting the equations for x ′ 0 x ′ 0 x^('0)x^{\prime 0}x′0 and x ′ 2 x ′ 2 x^('2)x^{\prime 2}x′2 into the equation for x ′ ′ 0 x ′ ′ 0 x^(''0)x^{\prime \prime 0}x′′0, we obtain
x ′ ′ 0 = x 0 cosh ⁡ θ cosh ⁡ θ ′ − x 1 sinh ⁡ θ cosh ⁡ θ ′ − x 2 sinh ⁡ θ ′ ≡ x 0 cosh ⁡ θ ′ ′ − ⋯ , x ′ ′ 0 = x 0 cosh ⁡ θ cosh ⁡ θ ′ − x 1 sinh ⁡ θ cosh ⁡ θ ′ − x 2 sinh ⁡ θ ′ ≡ x 0 cosh ⁡ θ ′ ′ − ⋯ , x^(''0)=x^(0)cosh theta cosh theta^(')-x^(1)sinh theta cosh theta^(')-x^(2)sinh theta^(')-=x^(0)cosh theta^('')-cdots,x^{\prime \prime 0}=x^{0} \cosh \theta \cosh \theta^{\prime}-x^{1} \sinh \theta \cosh \theta^{\prime}-x^{2} \sinh \theta^{\prime} \equiv x^{0} \cosh \theta^{\prime \prime}-\cdots,x′′0=x0cosh⁡θcosh⁡θ′−x1sinh⁡θcosh⁡θ′−x2sinh⁡θ′≡x0cosh⁡θ′′−⋯,
where tanh ⁡ θ ′ ′ ≡ v ′ ′ c tanh ⁡ θ ′ ′ ≡ v ′ ′ c tanh theta^('')-=(v^(''))/(c)\tanh \theta^{\prime \prime} \equiv \frac{v^{\prime \prime}}{c}tanh⁡θ′′≡v′′c. The velocity v ′ ′ v ′ ′ v^('')v^{\prime \prime}v′′ is the (magnitude of) the relative velocity between A A AAA and C C CCC. Using the hint, this means that
(3.118) cosh ⁡ θ ′ ′ = cosh ⁡ θ cosh ⁡ θ ′ (3.118) cosh ⁡ θ ′ ′ = cosh ⁡ θ cosh ⁡ θ ′ {:(3.118)cosh theta^('')=cosh theta cosh theta^('):}\begin{equation*} \cosh \theta^{\prime \prime}=\cosh \theta \cosh \theta^{\prime} \tag{3.118} \end{equation*}(3.118)cosh⁡θ′′=cosh⁡θcosh⁡θ′
i.e.,
(3.119) θ ′ ′ = arcosh ( cosh ⁡ θ cosh ⁡ θ ′ ) (3.119) θ ′ ′ = arcosh cosh ⁡ θ cosh ⁡ θ ′ {:(3.119)theta^('')=arcosh(cosh theta cosh theta^(')):}\begin{equation*} \theta^{\prime \prime}=\operatorname{arcosh}\left(\cosh \theta \cosh \theta^{\prime}\right) \tag{3.119} \end{equation*}(3.119)θ′′=arcosh(cosh⁡θcosh⁡θ′)
Thus, we have
(3.120) v ′ ′ = c tanh ⁡ θ ′ ′ = c tanh ⁡ arcosh ( cosh ⁡ θ cosh ⁡ θ ′ ) (3.120) v ′ ′ = c tanh ⁡ θ ′ ′ = c tanh ⁡ arcosh cosh ⁡ θ cosh ⁡ θ ′ {:(3.120)v^('')=c tanh theta^('')=c tanh arcosh(cosh theta cosh theta^(')):}\begin{equation*} v^{\prime \prime}=c \tanh \theta^{\prime \prime}=c \tanh \operatorname{arcosh}\left(\cosh \theta \cosh \theta^{\prime}\right) \tag{3.120} \end{equation*}(3.120)v′′=ctanh⁡θ′′=ctanh⁡arcosh(cosh⁡θcosh⁡θ′)
or with the rapidities inserted
(3.121) v ′ ′ = c tanh ⁡ arcosh ( cosh ⁡ artanh v c cosh ⁡ artanh v ′ c ) (3.121) v ′ ′ = c tanh ⁡ arcosh cosh ⁡ artanh v c cosh ⁡ artanh v ′ c {:(3.121)v^('')=c tanh arcosh(cosh artanh((v)/(c))cosh artanh((v^('))/(c))):}\begin{equation*} v^{\prime \prime}=c \tanh \operatorname{arcosh}\left(\cosh \operatorname{artanh} \frac{v}{c} \cosh \operatorname{artanh} \frac{v^{\prime}}{c}\right) \tag{3.121} \end{equation*}(3.121)v′′=ctanh⁡arcosh(cosh⁡artanhvccosh⁡artanhv′c)
We know that γ ( v ′ ′ ) = cosh ⁡ θ ′ ′ γ v ′ ′ = cosh ⁡ θ ′ ′ gamma(v^(''))=cosh theta^('')\gamma\left(v^{\prime \prime}\right)=\cosh \theta^{\prime \prime}γ(v′′)=cosh⁡θ′′. It follows that γ ( v ′ ′ ) = cosh ⁡ θ cosh ⁡ θ ′ = γ v ′ ′ = cosh ⁡ θ cosh ⁡ θ ′ = gamma(v^(''))=cosh theta cosh theta^(')=\gamma\left(v^{\prime \prime}\right)=\cosh \theta \cosh \theta^{\prime}=γ(v′′)=cosh⁡θcosh⁡θ′= γ ( v ) γ ( v ′ ) γ ( v ) γ v ′ gamma(v)gamma(v^('))\gamma(v) \gamma\left(v^{\prime}\right)γ(v)γ(v′), and thus, the time dilation formula between the time intervals Δ t ≡ Δ t ≡ Delta t-=\Delta t \equivΔt≡ t E 2 − t E 1 t E 2 − t E 1 t_(E_(2))-t_(E_(1))t_{E_{2}}-t_{E_{1}}tE2−tE1 and Δ t ′ ′ ≡ t E 2 ′ ′ − t E 1 ′ ′ Δ t ′ ′ ≡ t E 2 ′ ′ − t E 1 ′ ′ Deltat^('')-=t_(E_(2))^('')-t_(E_(1))^('')\Delta t^{\prime \prime} \equiv t_{E_{2}}^{\prime \prime}-t_{E_{1}}^{\prime \prime}Δt′′≡tE2′′−tE1′′ is given by
(3.122) Δ t ′ ′ = Δ t γ ( v ) γ ( v ′ ) = 1 − v 2 c 2 1 − v ′ 2 c 2 ⋅ Δ t (3.122) Δ t ′ ′ = Δ t γ ( v ) γ v ′ = 1 − v 2 c 2 1 − v ′ 2 c 2 ⋅ Δ t {:(3.122)Deltat^('')=(Delta t)/(gamma(v)gamma(v^(')))=sqrt(1-(v^(2))/(c^(2)))sqrt(1-(v^('2))/(c^(2)))*Delta t:}\begin{equation*} \Delta t^{\prime \prime}=\frac{\Delta t}{\gamma(v) \gamma\left(v^{\prime}\right)}=\sqrt{1-\frac{v^{2}}{c^{2}}} \sqrt{1-\frac{v^{\prime 2}}{c^{2}}} \cdot \Delta t \tag{3.122} \end{equation*}(3.122)Δt′′=Δtγ(v)γ(v′)=1−v2c21−v′2c2⋅Δt
1.35
Using
(3.123) u = N ( x 0 + x 3 x 1 + i x 2 ) ⇔ u ∗ = N ( x 0 + x 3 x 1 − i x 2 ) (3.123) u = N ( x 0 + x 3 x 1 + i x 2 ) ⇔ u ∗ = N x 0 + x 3 x 1 − i x 2 {:(3.123)u=N((x^(0)+x^(3))/(x^(1)+ix^(2)))quad<=>quadu^(**)=N(x^(0)+x^(3)quadx^(1)-ix^(2)):}\begin{equation*} u=N\binom{x^{0}+x^{3}}{x^{1}+i x^{2}} \quad \Leftrightarrow \quad u^{*}=N\left(x^{0}+x^{3} \quad x^{1}-i x^{2}\right) \tag{3.123} \end{equation*}(3.123)u=N(x0+x3x1+ix2)⇔u∗=N(x0+x3x1−ix2)
so we have
u u ∗ = N 2 ( ( x 0 + x 3 ) 2 ( x 0 + x 3 ) ( x 1 − i x 2 ) ( x 1 + i x 2 ) ( x 0 + x 3 ) ( x 1 ) 2 + ( x 2 ) 2 ) (3.124) = N 2 ( x 0 + x 3 ) ( x 0 + x 3 x 1 − i x 2 x 1 + i x 2 ( x 1 ) 2 + ( x 2 ) 2 x 0 + x 3 ) u u ∗ = N 2 x 0 + x 3 2 x 0 + x 3 x 1 − i x 2 x 1 + i x 2 x 0 + x 3 x 1 2 + x 2 2 (3.124) = N 2 x 0 + x 3 x 0 + x 3 x 1 − i x 2 x 1 + i x 2 x 1 2 + x 2 2 x 0 + x 3 {:[uu^(**)=N^(2)([(x^(0)+x^(3))^(2),(x^(0)+x^(3))(x^(1)-ix^(2))],[(x^(1)+ix^(2))(x^(0)+x^(3)),(x^(1))^(2)+(x^(2))^(2)])],[(3.124)=N^(2)(x^(0)+x^(3))([x^(0)+x^(3),x^(1)-ix^(2)],[x^(1)+ix^(2),((x^(1))^(2)+(x^(2))^(2))/(x^(0)+x^(3))])]:}\begin{align*} u u^{*} & =N^{2}\left(\begin{array}{cc} \left(x^{0}+x^{3}\right)^{2} & \left(x^{0}+x^{3}\right)\left(x^{1}-i x^{2}\right) \\ \left(x^{1}+i x^{2}\right)\left(x^{0}+x^{3}\right) & \left(x^{1}\right)^{2}+\left(x^{2}\right)^{2} \end{array}\right) \\ & =N^{2}\left(x^{0}+x^{3}\right)\left(\begin{array}{cc} x^{0}+x^{3} & x^{1}-i x^{2} \\ x^{1}+i x^{2} & \frac{\left(x^{1}\right)^{2}+\left(x^{2}\right)^{2}}{x^{0}+x^{3}} \end{array}\right) \tag{3.124} \end{align*}uu∗=N2((x0+x3)2(x0+x3)(x1−ix2)(x1+ix2)(x0+x3)(x1)2+(x2)2)(3.124)=N2(x0+x3)(x0+x3x1−ix2x1+ix2(x1)2+(x2)2x0+x3)
However, x 2 = 0 x 2 = 0 x^(2)=0x^{2}=0x2=0 implies that ( x 0 ) 2 = ( x 1 ) 2 + ( x 2 ) 2 + ( x 3 ) 2 x 0 2 = x 1 2 + x 2 2 + x 3 2 (x^(0))^(2)=(x^(1))^(2)+(x^(2))^(2)+(x^(3))^(2)\left(x^{0}\right)^{2}=\left(x^{1}\right)^{2}+\left(x^{2}\right)^{2}+\left(x^{3}\right)^{2}(x0)2=(x1)2+(x2)2+(x3)2, so we can write ( x 1 ) 2 + ( x 2 ) 2 = ( x 0 ) 2 − ( x 3 ) 2 = ( x 0 − x 3 ) ( x 0 + x 3 ) x 1 2 + x 2 2 = x 0 2 − x 3 2 = x 0 − x 3 x 0 + x 3 (x^(1))^(2)+(x^(2))^(2)=(x^(0))^(2)-(x^(3))^(2)=(x^(0)-x^(3))(x^(0)+x^(3))\left(x^{1}\right)^{2}+\left(x^{2}\right)^{2}=\left(x^{0}\right)^{2}-\left(x^{3}\right)^{2}=\left(x^{0}-x^{3}\right)\left(x^{0}+x^{3}\right)(x1)2+(x2)2=(x0)2−(x3)2=(x0−x3)(x0+x3). Therefore, we obtain
(3.125) u u ∗ = N 2 ( x 0 + x 3 ) ( x 0 + x 3 x 1 − i x 2 x 1 + i x 2 x 0 − x 3 ) (3.125) u u ∗ = N 2 x 0 + x 3 x 0 + x 3 x 1 − i x 2 x 1 + i x 2 x 0 − x 3 {:(3.125)uu^(**)=N^(2)(x^(0)+x^(3))([x^(0)+x^(3),x^(1)-ix^(2)],[x^(1)+ix^(2),x^(0)-x^(3)]):}u u^{*}=N^{2}\left(x^{0}+x^{3}\right)\left(\begin{array}{cc} x^{0}+x^{3} & x^{1}-i x^{2} \tag{3.125}\\ x^{1}+i x^{2} & x^{0}-x^{3} \end{array}\right)(3.125)uu∗=N2(x0+x3)(x0+x3x1−ix2x1+ix2x0−x3)
and we find that
(3.126) tr ( u u ∗ ) = N 2 ( x 0 + x 3 ) ⋅ 2 x 0 ≡ 2 x 0 ⇔ N = ± 1 x 0 + x 3 . (3.126) tr u u ∗ = N 2 x 0 + x 3 ⋅ 2 x 0 ≡ 2 x 0 ⇔ N = ± 1 x 0 + x 3 . {:(3.126)tr(uu^(**))=N^(2)(x^(0)+x^(3))*2x^(0)-=2x^(0)quad<=>quad N=+-(1)/(sqrt(x^(0)+x^(3))).:}\begin{equation*} \operatorname{tr}\left(u u^{*}\right)=N^{2}\left(x^{0}+x^{3}\right) \cdot 2 x^{0} \equiv 2 x^{0} \quad \Leftrightarrow \quad N= \pm \frac{1}{\sqrt{x^{0}+x^{3}}} . \tag{3.126} \end{equation*}(3.126)tr(uu∗)=N2(x0+x3)⋅2x0≡2x0⇔N=±1x0+x3.
Thus, it holds that
u = ( x 0 + x 3 x 0 + x 3 x 1 + i x 2 x 0 + x 3 ) is normalized (3.127) u u ∗ = ( x 0 + x 3 x 1 − i x 2 x 1 + i x 2 x 0 − x 3 ) u = ( x 0 + x 3 x 0 + x 3 x 1 + i x 2 x 0 + x 3 )  is normalized  (3.127) u u ∗ = x 0 + x 3 x 1 − i x 2 x 1 + i x 2 x 0 − x 3 {:[u=(((x^(0)+x^(3))/(sqrt(x^(0)+x^(3))))/((x^(1)+ix^(2))/(sqrt(x^(0)+x^(3)))))quad" is normalized "],[(3.127)uu^(**)=([x^(0)+x^(3),x^(1)-ix^(2)],[x^(1)+ix^(2),x^(0)-x^(3)])]:}\begin{align*} u & =\binom{\frac{x^{0}+x^{3}}{\sqrt{x^{0}+x^{3}}}}{\frac{x^{1}+i x^{2}}{\sqrt{x^{0}+x^{3}}}} \quad \text { is normalized } \\ u u^{*} & =\left(\begin{array}{cc} x^{0}+x^{3} & x^{1}-i x^{2} \\ x^{1}+i x^{2} & x^{0}-x^{3} \end{array}\right) \tag{3.127} \end{align*}u=(x0+x3x0+x3x1+ix2x0+x3) is normalized (3.127)uu∗=(x0+x3x1−ix2x1+ix2x0−x3)
and
(3.128) det ( u u ∗ ) = ( x 0 ) 2 − ( x 1 ) 2 − ( x 2 ) 2 − ( x 3 ) 2 = 0 (3.128) det u u ∗ = x 0 2 − x 1 2 − x 2 2 − x 3 2 = 0 {:(3.128)det(uu^(**))=(x^(0))^(2)-(x^(1))^(2)-(x^(2))^(2)-(x^(3))^(2)=0:}\begin{equation*} \operatorname{det}\left(u u^{*}\right)=\left(x^{0}\right)^{2}-\left(x^{1}\right)^{2}-\left(x^{2}\right)^{2}-\left(x^{3}\right)^{2}=0 \tag{3.128} \end{equation*}(3.128)det(uu∗)=(x0)2−(x1)2−(x2)2−(x3)2=0
Consider the Lorentz transformation of u u uuu, i.e.,
(3.129) a ( v ) u = ( e − θ / 2 0 0 e θ / 2 ) ( x 0 + x 3 x 0 + x 3 x 1 + i x 2 x 0 + x 3 ) = ( e − θ / 2 x 0 + x 3 x 0 + x 3 e θ / 2 x 1 + i x 2 x 0 + x 3 ) = ( e − θ ( x 0 + x 3 ) e − θ ( x 0 + x 3 ) x 1 + i x 2 e − θ ( x 0 + x 3 ) ) (3.129) a ( v ) u = e − θ / 2 0 0 e θ / 2 ( x 0 + x 3 x 0 + x 3 x 1 + i x 2 x 0 + x 3 ) = ( e − θ / 2 x 0 + x 3 x 0 + x 3 e θ / 2 x 1 + i x 2 x 0 + x 3 ) = ( e − θ x 0 + x 3 e − θ x 0 + x 3 x 1 + i x 2 e − θ x 0 + x 3 ) {:(3.129)a(v)u=([e^(-theta//2),0],[0,e^(theta//2)])(((x^(0)+x^(3))/(sqrt(x^(0)+x^(3))))/((x^(1)+ix^(2))/(sqrt(x^(0)+x^(3)))))=((e^(-theta//2)(x^(0)+x^(3))/(sqrt(x^(0)+x^(3))))/(e^(theta//2)(x^(1)+ix^(2))/(sqrt(x^(0)+x^(3)))))=(((e^(-theta)(x^(0)+x^(3)))/(sqrt(e^(-theta)(x^(0)+x^(3)))))/((x^(1)+ix^(2))/(sqrt(e^(-theta)(x^(0)+x^(3)))))):}a(v) u=\left(\begin{array}{cc} e^{-\theta / 2} & 0 \tag{3.129}\\ 0 & e^{\theta / 2} \end{array}\right)\binom{\frac{x^{0}+x^{3}}{\sqrt{x^{0}+x^{3}}}}{\frac{x^{1}+i x^{2}}{\sqrt{x^{0}+x^{3}}}}=\binom{e^{-\theta / 2} \frac{x^{0}+x^{3}}{\sqrt{x^{0}+x^{3}}}}{e^{\theta / 2} \frac{x^{1}+i x^{2}}{\sqrt{x^{0}+x^{3}}}}=\binom{\frac{e^{-\theta}\left(x^{0}+x^{3}\right)}{\sqrt{e^{-\theta}\left(x^{0}+x^{3}\right)}}}{\frac{x^{1}+i x^{2}}{\sqrt{e^{-\theta}\left(x^{0}+x^{3}\right)}}}(3.129)a(v)u=(e−θ/200eθ/2)(x0+x3x0+x3x1+ix2x0+x3)=(e−θ/2x0+x3x0+x3eθ/2x1+ix2x0+x3)=(e−θ(x0+x3)e−θ(x0+x3)x1+ix2e−θ(x0+x3))
Now, the Lorentz transformation of x x xxx is given by
(3.130) { x ′ 0 = x 0 cosh ⁡ θ − x 3 sinh ⁡ θ x ′ 1 = x 1 x ′ 2 = x 2 x ′ 3 = − x 0 sinh ⁡ θ + x 3 cosh ⁡ θ (3.130) x ′ 0 = x 0 cosh ⁡ θ − x 3 sinh ⁡ θ x ′ 1 = x 1 x ′ 2 = x 2 x ′ 3 = − x 0 sinh ⁡ θ + x 3 cosh ⁡ θ {:(3.130){[x^('0)=x^(0)cosh theta-x^(3)sinh theta],[x^('1)=x^(1)],[x^('2)=x^(2)],[x^('3)=-x^(0)sinh theta+x^(3)cosh theta]:}:}\left\{\begin{array}{l} x^{\prime 0}=x^{0} \cosh \theta-x^{3} \sinh \theta \tag{3.130}\\ x^{\prime 1}=x^{1} \\ x^{\prime 2}=x^{2} \\ x^{\prime 3}=-x^{0} \sinh \theta+x^{3} \cosh \theta \end{array}\right.(3.130){x′0=x0cosh⁡θ−x3sinh⁡θx′1=x1x′2=x2x′3=−x0sinh⁡θ+x3cosh⁡θ
Therefore, we find that
x ′ 0 + x ′ 3 = x 0 ( cosh ⁡ θ − sinh ⁡ θ ) + x 3 ( − sinh ⁡ θ + cosh ⁡ θ ) (3.131) = ( cosh ⁡ θ − sinh ⁡ θ ) ( x 0 + x 3 ) x ′ 0 + x ′ 3 = x 0 ( cosh ⁡ θ − sinh ⁡ θ ) + x 3 ( − sinh ⁡ θ + cosh ⁡ θ ) (3.131) = ( cosh ⁡ θ − sinh ⁡ θ ) x 0 + x 3 {:[x^('0)+x^('3)=x^(0)(cosh theta-sinh theta)+x^(3)(-sinh theta+cosh theta)],[(3.131)=(cosh theta-sinh theta)(x^(0)+x^(3))]:}\begin{align*} x^{\prime 0}+x^{\prime 3} & =x^{0}(\cosh \theta-\sinh \theta)+x^{3}(-\sinh \theta+\cosh \theta) \\ & =(\cosh \theta-\sinh \theta)\left(x^{0}+x^{3}\right) \tag{3.131} \end{align*}x′0+x′3=x0(cosh⁡θ−sinh⁡θ)+x3(−sinh⁡θ+cosh⁡θ)(3.131)=(cosh⁡θ−sinh⁡θ)(x0+x3)
and so we have
(3.132) cosh ⁡ θ − sinh ⁡ θ = e θ + e − θ 2 − e θ − e − θ 2 = e − θ (3.132) cosh ⁡ θ − sinh ⁡ θ = e θ + e − θ 2 − e θ − e − θ 2 = e − θ {:(3.132)cosh theta-sinh theta=(e^(theta)+e^(-theta))/(2)-(e^(theta)-e^(-theta))/(2)=e^(-theta):}\begin{equation*} \cosh \theta-\sinh \theta=\frac{e^{\theta}+e^{-\theta}}{2}-\frac{e^{\theta}-e^{-\theta}}{2}=e^{-\theta} \tag{3.132} \end{equation*}(3.132)cosh⁡θ−sinh⁡θ=eθ+e−θ2−eθ−e−θ2=e−θ
Thus, we obtain
(3.133) a ( v ) u = ( x ′ 0 + x ′ 3 x ′ 0 + x ′ 3 x ′ 1 + i x ′ 2 x ′ 0 + x ′ 3 ) = u ( x ′ ) = u ( L ( a ( v ) ) x ) (3.133) a ( v ) u = ( x ′ 0 + x ′ 3 x ′ 0 + x ′ 3 x ′ 1 + i x ′ 2 x ′ 0 + x ′ 3 ) = u x ′ = u ( L ( a ( v ) ) x ) {:(3.133)a(v)u=(((x^('0)+x^('3))/(sqrt(x^('0)+x^('3))))/((x^('1)+ix^('2))/(sqrt(x^('0)+x^('3)))))=u(x^('))=u(L(a(v))x):}\begin{equation*} a(v) u=\binom{\frac{x^{\prime 0}+x^{\prime 3}}{\sqrt{x^{\prime 0}+x^{\prime 3}}}}{\frac{x^{\prime 1}+i x^{\prime 2}}{\sqrt{x^{\prime 0}+x^{\prime 3}}}}=u\left(x^{\prime}\right)=u(L(a(v)) x) \tag{3.133} \end{equation*}(3.133)a(v)u=(x′0+x′3x′0+x′3x′1+ix′2x′0+x′3)=u(x′)=u(L(a(v))x)
1.36
A Lorentz transformation is linear and we will derive the expressions for a boost in the x x xxx-direction. The general linear transformation is given by
(3.134) x ′ = γ x + b t (3.135) t ′ = A x + B t (3.134) x ′ = γ x + b t (3.135) t ′ = A x + B t {:[(3.134)x^(')=gamma x+bt],[(3.135)t^(')=Ax+Bt]:}\begin{align*} x^{\prime} & =\gamma x+b t \tag{3.134}\\ t^{\prime} & =A x+B t \tag{3.135} \end{align*}(3.134)x′=γx+bt(3.135)t′=Ax+Bt
for a transformation from an inertial frame S S SSS to an inertial frame S ′ S ′ S^(')S^{\prime}S′ moving with a speed v v vvv relative to S S SSS. In the case that x = v t x = v t x=vtx=v tx=vt, i.e., moving along with S ′ S ′ S^(')S^{\prime}S′, we must have x ′ = 0 x ′ = 0 x^(')=0x^{\prime}=0x′=0. Hence, we have
(3.136) x ′ = γ v t + b t = ( γ v + b ) t = 0 ⇒ b = − γ v (3.136) x ′ = γ v t + b t = ( γ v + b ) t = 0 ⇒ b = − γ v {:(3.136)x^(')=gamma vt+bt=(gamma v+b)t=0quad=>quad b=-gamma v:}\begin{equation*} x^{\prime}=\gamma v t+b t=(\gamma v+b) t=0 \quad \Rightarrow \quad b=-\gamma v \tag{3.136} \end{equation*}(3.136)x′=γvt+bt=(γv+b)t=0⇒b=−γv
and thus, we obtain
(3.137) x ′ = γ ( x − v t ) , (3.138) x = γ ( x ′ + v t ′ ) (3.137) x ′ = γ ( x − v t ) , (3.138) x = γ x ′ + v t ′ {:[(3.137)x^(')=gamma(x-vt)","],[(3.138)x=gamma(x^(')+vt^('))]:}\begin{align*} x^{\prime} & =\gamma(x-v t), \tag{3.137}\\ x & =\gamma\left(x^{\prime}+v t^{\prime}\right) \tag{3.138} \end{align*}(3.137)x′=γ(x−vt),(3.138)x=γ(x′+vt′)
where for the second equation we have assumed that the situation must be symmetric to the first one. Furthermore, we must have t = x / c t = x / c t=x//ct=x / ct=x/c and t ′ = x ′ / c t ′ = x ′ / c t^(')=x^(')//ct^{\prime}=x^{\prime} / ct′=x′/c for a light ray. Therefore, we find that
(3.139) x ′ = γ ( 1 − v c ) x (3.140) x = γ ( 1 + v c ) x ′ (3.139) x ′ = γ 1 − v c x (3.140) x = γ 1 + v c x ′ {:[(3.139)x^(')=gamma(1-(v)/(c))x],[(3.140)x=gamma(1+(v)/(c))x^(')]:}\begin{align*} x^{\prime} & =\gamma\left(1-\frac{v}{c}\right) x \tag{3.139}\\ x & =\gamma\left(1+\frac{v}{c}\right) x^{\prime} \tag{3.140} \end{align*}(3.139)x′=γ(1−vc)x(3.140)x=γ(1+vc)x′
Thus, inserting the first equation into the second one, we obtain
(3.141) x = γ 2 ( 1 − v 2 c 2 ) x ⇒ γ = 1 1 − v 2 c 2 (3.141) x = γ 2 1 − v 2 c 2 x ⇒ γ = 1 1 − v 2 c 2 {:(3.141)x=gamma^(2)(1-(v^(2))/(c^(2)))x=>gamma=(1)/(sqrt(1-(v^(2))/(c^(2)))):}\begin{equation*} x=\gamma^{2}\left(1-\frac{v^{2}}{c^{2}}\right) x \Rightarrow \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \tag{3.141} \end{equation*}(3.141)x=γ2(1−v2c2)x⇒γ=11−v2c2
In addition, we can determine the parameters A A AAA and B B BBB to be A = − γ v c 2 A = − γ v c 2 A=-gamma(v)/(c^(2))A=-\gamma \frac{v}{c^{2}}A=−γvc2 and B = γ B = γ B=gammaB=\gammaB=γ by requiring the inverse transformation to take a similar form.

1.37

Calling the events where the light signals reach the rockets 1 and 2 , respectively (with 2 being the event for the rocket traveling with velocity v v vvv ), event 2 is given by the intersection of the worldlines
(3.142) x = v t and x = t − t 0 (3.142) x = v t  and  x = t − t 0 {:(3.142)x=vt quad" and "quad x=t-t_(0):}\begin{equation*} x=v t \quad \text { and } \quad x=t-t_{0} \tag{3.142} \end{equation*}(3.142)x=vt and x=t−t0
This leads to
(3.143) t 2 = t 0 1 − v and x 2 = v t 2 = v t 0 1 − v (3.143) t 2 = t 0 1 − v  and  x 2 = v t 2 = v t 0 1 − v {:(3.143)t_(2)=(t_(0))/(1-v)quad" and "quadx_(2)=vt_(2)=(vt_(0))/(1-v):}\begin{equation*} t_{2}=\frac{t_{0}}{1-v} \quad \text { and } \quad x_{2}=v t_{2}=\frac{v t_{0}}{1-v} \tag{3.143} \end{equation*}(3.143)t2=t01−v and x2=vt2=vt01−v
By symmetry, t 1 = t 2 t 1 = t 2 t_(1)=t_(2)t_{1}=t_{2}t1=t2 and x 1 = − x 2 x 1 = − x 2 x_(1)=-x_(2)x_{1}=-x_{2}x1=−x2. It follows that the spacetime separation between the events is given by
(3.144) Δ x = x 2 − x 1 = 2 v t 0 1 − v and Δ t = t 2 − t 1 = 0 (3.144) Δ x = x 2 − x 1 = 2 v t 0 1 − v  and  Δ t = t 2 − t 1 = 0 {:(3.144)Delta x=x_(2)-x_(1)=(2vt_(0))/(1-v)quad" and "quad Delta t=t_(2)-t_(1)=0:}\begin{equation*} \Delta x=x_{2}-x_{1}=\frac{2 v t_{0}}{1-v} \quad \text { and } \quad \Delta t=t_{2}-t_{1}=0 \tag{3.144} \end{equation*}(3.144)Δx=x2−x1=2vt01−v and Δt=t2−t1=0
Lorentz transforming this to the rest frame S ′ S ′ S^(')S^{\prime}S′ of rocket 2, we find that
(3.145) Δ t ′ = t 2 ′ − t 1 ′ = γ ( Δ t − v Δ x ) = − 2 v 2 γ t 0 1 − v (3.145) Δ t ′ = t 2 ′ − t 1 ′ = γ ( Δ t − v Δ x ) = − 2 v 2 γ t 0 1 − v {:(3.145)Deltat^(')=t_(2)^(')-t_(1)^(')=gamma(Delta t-v Delta x)=-(2v^(2)gammat_(0))/(1-v):}\begin{equation*} \Delta t^{\prime}=t_{2}^{\prime}-t_{1}^{\prime}=\gamma(\Delta t-v \Delta x)=-\frac{2 v^{2} \gamma t_{0}}{1-v} \tag{3.145} \end{equation*}(3.145)Δt′=t2′−t1′=γ(Δt−vΔx)=−2v2γt01−v
Thus, in the rest frame of one of the rockets, the other rocket receives the signal a time
(3.146) − Δ t ′ = 2 v 2 γ t 0 1 − v (3.146) − Δ t ′ = 2 v 2 γ t 0 1 − v {:(3.146)-Deltat^(')=(2v^(2)gammat_(0))/(1-v):}\begin{equation*} -\Delta t^{\prime}=\frac{2 v^{2} \gamma t_{0}}{1-v} \tag{3.146} \end{equation*}(3.146)−Δt′=2v2γt01−v
later.

1.38

a) Relativity of simultaneity means that it is possible for two events that are simultaneous in one inertial frame to not be simultaneous in a different frame. This occurs if there is a spatial separation between the events in the direction of the relative velocity between the frames.
The concept of relativity of simultaneity can be illustrated by the spacetime diagram shown in Figure 3.6. In this spacetime diagram, there are two events A and B , where A occurs at the origin of both frames O O O\mathcal{O}O and O ′ O ′ O^(')\mathcal{O}^{\prime}O′, i.e., x A = x A ′ = 0 x A = x A ′ = 0 x_(A)=x_(A)^(')=0x_{\mathrm{A}}=x_{\mathrm{A}}^{\prime}=0xA=xA′=0, at equal times t A = t A ′ = 0 t A = t A ′ = 0 t_(A)=t_(A)^(')=0t_{\mathrm{A}}=t_{\mathrm{A}}^{\prime}=0tA=tA′=0, whereas B occurs at x B x B x_(B)x_{\mathrm{B}}xB in O O O\mathcal{O}O at time t B = t A = 0 t B = t A = 0 t_(B)=t_(A)=0t_{\mathrm{B}}=t_{\mathrm{A}}=0tB=tA=0, but at x B ′ x B ′ x_(B)^(')x_{\mathrm{B}}^{\prime}xB′
Figure 3.6 Relativity of simultaneity: t A = t B t A = t B t_(A)=t_(B)t_{\mathrm{A}}=t_{\mathrm{B}}tA=tB, but t A ′ > t B ′ t A ′ > t B ′ t_(A)^(') > t_(B)^(')t_{\mathrm{A}}^{\prime}>t_{\mathrm{B}}^{\prime}tA′>tB′.
in O ′ O ′ O^(')\mathcal{O}^{\prime}O′ at time t B ′ ≠ t A ′ t B ′ ≠ t A ′ t_(B)^(')!=t_(A)^(')t_{\mathrm{B}}^{\prime} \neq t_{\mathrm{A}}^{\prime}tB′≠tA′. In fact, the two events are simultaneous in O O O\mathcal{O}O, since t A = t B t A = t B t_(A)=t_(B)t_{\mathrm{A}}=t_{\mathrm{B}}tA=tB, but they are not simultaneous in O ′ O ′ O^(')\mathcal{O}^{\prime}O′, since t A ′ > t B ′ t A ′ > t B ′ t_(A)^(') > t_(B)^(')t_{\mathrm{A}}^{\prime}>t_{\mathrm{B}}^{\prime}tA′>tB′. In mathematical language, using the Lorentz transformation between O O O\mathcal{O}O and O ′ O ′ O^(')\mathcal{O}^{\prime}O′ (assuming that the frames are moving with constant speed v v vvv relative to each other), we obtain the relations for B as x B ′ = γ ( v ) x B x B ′ = γ ( v ) x B x_(B)^(')=gamma(v)x_(B)x_{\mathrm{B}}^{\prime}=\gamma(v) x_{\mathrm{B}}xB′=γ(v)xB and t B ′ = − γ ( v ) v x B / c 2 t B ′ = − γ ( v ) v x B / c 2 t_(B)^(')=-gamma(v)vx_(B)//c^(2)t_{\mathrm{B}}^{\prime}=-\gamma(v) v x_{\mathrm{B}} / c^{2}tB′=−γ(v)vxB/c2, where γ ( v ) ≡ 1 / 1 − v 2 / c 2 γ ( v ) ≡ 1 / 1 − v 2 / c 2 gamma(v)-=1//sqrt(1-v^(2)//c^(2))\gamma(v) \equiv 1 / \sqrt{1-v^{2} / c^{2}}γ(v)≡1/1−v2/c2. Note that x B ′ ≥ x B x B ′ ≥ x B x_(B)^(') >= x_(B)x_{\mathrm{B}}^{\prime} \geq x_{\mathrm{B}}xB′≥xB, since γ ( v ) ≥ 1 γ ( v ) ≥ 1 gamma(v) >= 1\gamma(v) \geq 1γ(v)≥1, and t B ′ ≤ 0 t B ′ ≤ 0 t_(B)^(') <= 0t_{\mathrm{B}}^{\prime} \leq 0tB′≤0 and directly proportional to the spatial coordinate x B x B x_(B)x_{\mathrm{B}}xB.
b) We compute
(3.147) v = ( d x d t , d y d t , d z d t ) = a t ( 3 , 4 , 0 ) (3.147) v = d x d t , d y d t , d z d t = a t ( 3 , 4 , 0 ) {:(3.147)v=((dx)/(dt),(dy)/(dt),(dz)/(dt))=at(3","4","0):}\begin{equation*} v=\left(\frac{d x}{d t}, \frac{d y}{d t}, \frac{d z}{d t}\right)=a t(3,4,0) \tag{3.147} \end{equation*}(3.147)v=(dxdt,dydt,dzdt)=at(3,4,0)
and thus, we find that v 2 = ( 5 a t ) 2 v 2 = ( 5 a t ) 2 v^(2)=(5at)^(2)\boldsymbol{v}^{2}=(5 a t)^{2}v2=(5at)2. The condition | v | < c | v | < c |v| < c|\boldsymbol{v}|<c|v|<c implies t 0 < c / ( 5 a ) t 0 < c / ( 5 a ) t_(0) < c//(5a)t_{0}<c /(5 a)t0<c/(5a). We also find
(3.148) γ = 1 1 − ( v / c ) 2 = 1 1 − ( 5 a t / c ) 2 (3.148) γ = 1 1 − ( v / c ) 2 = 1 1 − ( 5 a t / c ) 2 {:(3.148)gamma=(1)/(sqrt(1-(v//c)^(2)))=(1)/(sqrt(1-(5at//c)^(2))):}\begin{equation*} \gamma=\frac{1}{\sqrt{1-(v / c)^{2}}}=\frac{1}{\sqrt{1-(5 a t / c)^{2}}} \tag{3.148} \end{equation*}(3.148)γ=11−(v/c)2=11−(5at/c)2
Thus, we can compute the 4 -velocity and the 4 -acceleration to be
(3.149) ( V μ ) = ( c γ , γ d x d t , γ d y d t , γ d z d t ) = γ ( c , 3 a t , 4 a t , 0 ) (3.150) ( A μ ) = γ d d t ( V μ ) = γ ( c d γ d t , 3 a d ( γ t ) d t , 4 a d ( γ t ) d t , 0 ) = a γ 4 ( 25 a t / c , 3 , 4 , 0 ) (3.149) V μ = c γ , γ d x d t , γ d y d t , γ d z d t = γ ( c , 3 a t , 4 a t , 0 ) (3.150) A μ = γ d d t V μ = γ c d γ d t , 3 a d ( γ t ) d t , 4 a d ( γ t ) d t , 0 = a γ 4 ( 25 a t / c , 3 , 4 , 0 ) {:[(3.149)(V^(mu))=(c gamma,gamma(dx)/(dt),gamma(dy)/(dt),gamma(dz)/(dt))=gamma(c","3at","4at","0)],[(3.150)(A^(mu))=gamma(d)/(dt)(V^(mu))=gamma(c(d gamma)/(dt),3a(d(gamma t))/(dt),4a(d(gamma t))/(dt),0)=agamma^(4)(25 at//c","3","4","0)]:}\begin{align*} & \left(V^{\mu}\right)=\left(c \gamma, \gamma \frac{d x}{d t}, \gamma \frac{d y}{d t}, \gamma \frac{d z}{d t}\right)=\gamma(c, 3 a t, 4 a t, 0) \tag{3.149}\\ & \left(A^{\mu}\right)=\gamma \frac{d}{d t}\left(V^{\mu}\right)=\gamma\left(c \frac{d \gamma}{d t}, 3 a \frac{d(\gamma t)}{d t}, 4 a \frac{d(\gamma t)}{d t}, 0\right)=a \gamma^{4}(25 a t / c, 3,4,0) \tag{3.150} \end{align*}(3.149)(Vμ)=(cγ,γdxdt,γdydt,γdzdt)=γ(c,3at,4at,0)(3.150)(Aμ)=γddt(Vμ)=γ(cdγdt,3ad(γt)dt,4ad(γt)dt,0)=aγ4(25at/c,3,4,0)
respectively. The proper time along the trajectory is
τ = ∫ 0 t 0 1 − ( v ( t ) / c ) 2 d t = ∫ 0 t 0 1 − ( 5 a t / c ) 2 d t (3.151) = t 0 2 1 − ( 5 a t 0 / c ) 2 + c 10 a arcsin ⁡ ( 5 a t 0 / c ) τ = ∫ 0 t 0   1 − ( v ( t ) / c ) 2 d t = ∫ 0 t 0   1 − ( 5 a t / c ) 2 d t (3.151) = t 0 2 1 − 5 a t 0 / c 2 + c 10 a arcsin ⁡ 5 a t 0 / c {:[tau=int_(0)^(t_(0))sqrt(1-(v(t)//c)^(2))dt=int_(0)^(t_(0))sqrt(1-(5at//c)^(2))dt],[(3.151)=(t_(0))/(2)sqrt(1-(5at_(0)//c)^(2))+(c)/(10 a)arcsin(5at_(0)//c)]:}\begin{align*} \tau & =\int_{0}^{t_{0}} \sqrt{1-(\boldsymbol{v}(t) / c)^{2}} d t=\int_{0}^{t_{0}} \sqrt{1-(5 a t / c)^{2}} d t \\ & =\frac{t_{0}}{2} \sqrt{1-\left(5 a t_{0} / c\right)^{2}}+\frac{c}{10 a} \arcsin \left(5 a t_{0} / c\right) \tag{3.151} \end{align*}τ=∫0t01−(v(t)/c)2dt=∫0t01−(5at/c)2dt(3.151)=t021−(5at0/c)2+c10aarcsin⁡(5at0/c)

1.39

Solution 1: Denote the inertial frame of the observer who measures the length of the rod to be L L LLL by S S SSS, the rest frame of the other observer by S ′ S ′ S^(')S^{\prime}S′, and the rest frame of the rod by S ′ ′ S ′ ′ S^('')S^{\prime \prime}S′′. By the Lorentz contraction formula, the length of the rod in S ′ ′ S ′ ′ S^('')S^{\prime \prime}S′′ is given by
(3.152) L ′ ′ = γ ( v ) L (3.152) L ′ ′ = γ ( v ) L {:(3.152)L^('')=gamma(v)L:}\begin{equation*} L^{\prime \prime}=\gamma(v) L \tag{3.152} \end{equation*}(3.152)L′′=γ(v)L
The relative velocity v ′ v ′ v^(')v^{\prime}v′ between S ′ S ′ S^(')S^{\prime}S′ and S ′ ′ S ′ ′ S^('')S^{\prime \prime}S′′ is given by relativistic addition of velocities
(3.153) v ′ = 2 v 1 + v 2 (3.153) v ′ = 2 v 1 + v 2 {:(3.153)v^(')=(2v)/(1+v^(2)):}\begin{equation*} v^{\prime}=\frac{2 v}{1+v^{2}} \tag{3.153} \end{equation*}(3.153)v′=2v1+v2
By the Lorentz contraction formula, the length of the rod in S ′ S ′ S^(')S^{\prime}S′ is given by
(3.154) L ′ = L ′ ′ γ ( v ′ ) = γ ( v ) L γ ( v ′ ) = 1 − v 2 1 + v 2 ⋅ L (3.154) L ′ = L ′ ′ γ v ′ = γ ( v ) L γ v ′ = 1 − v 2 1 + v 2 ⋅ L {:(3.154)L^(')=(L^(''))/(gamma(v^(')))=(gamma(v)L)/(gamma(v^(')))=(sqrt(1-v^(2)))/(1+v^(2))*L:}\begin{equation*} L^{\prime}=\frac{L^{\prime \prime}}{\gamma\left(v^{\prime}\right)}=\frac{\gamma(v) L}{\gamma\left(v^{\prime}\right)}=\frac{\sqrt{1-v^{2}}}{1+v^{2}} \cdot L \tag{3.154} \end{equation*}(3.154)L′=L′′γ(v′)=γ(v)Lγ(v′)=1−v21+v2⋅L
Solution 2: Let x A ( t A ) = ( t A , v t A ) x A t A = t A , v t A x_(A)(t_(A))=(t_(A),vt_(A))x_{A}\left(t_{A}\right)=\left(t_{A}, v t_{A}\right)xA(tA)=(tA,vtA) and x B ( t B ) = ( t B , v t B + L ) x B t B = t B , v t B + L x_(B)(t_(B))=(t_(B),vt_(B)+L)x_{B}\left(t_{B}\right)=\left(t_{B}, v t_{B}+L\right)xB(tB)=(tB,vtB+L) be the worldlines of the ends of the rod. Also, let x A ′ x A ′ x_(A)^(')x_{A}^{\prime}xA′ and x B ′ x B ′ x_(B)^(')x_{B}^{\prime}xB′ be events on those worldlines which are simultaneous in S ′ S ′ S^(')S^{\prime}S′. Without loss of generality, we can choose x A ′ = ( 0 , 0 ) x A ′ = ( 0 , 0 ) x_(A)^(')=(0,0)x_{A}^{\prime}=(0,0)xA′=(0,0). By Lorentz transformation, we find that the worldline x B x B x_(B)x_{B}xB is given by
(3.155) x B ′ ( t B ) = γ ( v ) ( t B ( 1 + v 2 ) + L v , 2 v t B + L ) (3.155) x B ′ t B = γ ( v ) t B 1 + v 2 + L v , 2 v t B + L {:(3.155)x_(B)^(')(t_(B))=gamma(v)(t_(B)(1+v^(2))+Lv,2vt_(B)+L):}\begin{equation*} x_{B}^{\prime}\left(t_{B}\right)=\gamma(v)\left(t_{B}\left(1+v^{2}\right)+L v, 2 v t_{B}+L\right) \tag{3.155} \end{equation*}(3.155)xB′(tB)=γ(v)(tB(1+v2)+Lv,2vtB+L)
in the S ′ S ′ S^(')S^{\prime}S′ frame. Using that x A ′ x A ′ x_(A)^(')x_{A}^{\prime}xA′ and x B ′ x B ′ x_(B)^(')x_{B}^{\prime}xB′ are simultaneous in S ′ S ′ S^(')S^{\prime}S′, we obtain that x B ′ = x B ′ = x_(B)^(')=x_{B}^{\prime}=xB′= x B ′ ( τ ) x B ′ ( τ ) x_(B)^(')(tau)x_{B}^{\prime}(\tau)xB′(τ) where τ = − L v / ( 1 + v 2 ) τ = − L v / 1 + v 2 tau=-Lv//(1+v^(2))\tau=-L v /\left(1+v^{2}\right)τ=−Lv/(1+v2). Thus, after simplification, the expression for the length of the rod for an observer at rest in S ′ S ′ S^(')S^{\prime}S′ is given by
(3.156) L ′ = x B ′ 1 − x A ′ 1 = 1 − v 2 1 + v 2 L (3.156) L ′ = x B ′ 1 − x A ′ 1 = 1 − v 2 1 + v 2 L {:(3.156)L^(')=x_(B)^(')^(1)-x_(A)^('1)=(sqrt(1-v^(2)))/(1+v^(2))L:}\begin{equation*} L^{\prime}=x_{B}^{\prime}{ }^{1}-x_{A}^{\prime 1}=\frac{\sqrt{1-v^{2}}}{1+v^{2}} L \tag{3.156} \end{equation*}(3.156)L′=xB′1−xA′1=1−v21+v2L

1.40

The 4-velocity of the particle in S S SSS can be written as
(3.157) V μ = d x μ d s = γ ( v ) ( 1 , v ) = γ ( v ) ( 1 , v 1 , v 2 , v 3 ) (3.157) V μ = d x μ d s = γ ( v ) ( 1 , v ) = γ ( v ) 1 , v 1 , v 2 , v 3 {:(3.157)V^(mu)=(dx^(mu))/(ds)=gamma(v)(1","v)=gamma(v)(1,v^(1),v^(2),v^(3)):}\begin{equation*} V^{\mu}=\frac{d x^{\mu}}{d s}=\gamma(\boldsymbol{v})(1, \boldsymbol{v})=\gamma(\boldsymbol{v})\left(1, v^{1}, v^{2}, v^{3}\right) \tag{3.157} \end{equation*}(3.157)Vμ=dxμds=γ(v)(1,v)=γ(v)(1,v1,v2,v3)
This is related to the 4-velocity in S ′ S ′ S^(')S^{\prime}S′ by the Lorentz transformation
(3.158) V ′ μ = Λ v μ V v , (3.158) V ′ μ = Λ v μ V v , {:(3.158)V^('mu)=Lambda_(v)^(mu)V^(v)",":}\begin{equation*} V^{\prime \mu}=\Lambda_{v}^{\mu} V^{v}, \tag{3.158} \end{equation*}(3.158)V′μ=ΛvμVv,
where
(3.159) ( Λ v μ ) = ( γ ( u ) 0 − u γ ( u ) 0 0 1 0 0 − u γ ( u ) 0 γ ( u ) 0 0 0 0 1 ) (3.159) Λ v μ = γ ( u ) 0 − u γ ( u ) 0 0 1 0 0 − u γ ( u ) 0 γ ( u ) 0 0 0 0 1 {:(3.159)(Lambda_(v)^(mu))=([gamma(u),0,-u gamma(u),0],[0,1,0,0],[-u gamma(u),0,gamma(u),0],[0,0,0,1]):}\left(\Lambda_{v}^{\mu}\right)=\left(\begin{array}{cccc} \gamma(u) & 0 & -u \gamma(u) & 0 \tag{3.159}\\ 0 & 1 & 0 & 0 \\ -u \gamma(u) & 0 & \gamma(u) & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)(3.159)(Λvμ)=(γ(u)0−uγ(u)00100−uγ(u)0γ(u)00001)
Thus, we find that
(3.160) V ′ μ = γ ( v ) ( γ ( u ) ( 1 − u v 2 ) , v 1 , γ ( u ) ( v 2 − u ) , v 3 ) (3.160) V ′ μ = γ ( v ) γ ( u ) 1 − u v 2 , v 1 , γ ( u ) v 2 − u , v 3 {:(3.160)V^('mu)=gamma(v)(gamma(u)(1-uv^(2)),v^(1),gamma(u)(v^(2)-u),v^(3)):}\begin{equation*} V^{\prime \mu}=\gamma(v)\left(\gamma(u)\left(1-u v^{2}\right), v^{1}, \gamma(u)\left(v^{2}-u\right), v^{3}\right) \tag{3.160} \end{equation*}(3.160)V′μ=γ(v)(γ(u)(1−uv2),v1,γ(u)(v2−u),v3)
By using the relation v ′ = V ′ / V ′ 0 v ′ = V ′ / V ′ 0 v^(')=V^(')//V^('0)\boldsymbol{v}^{\prime}=V^{\prime} / V^{\prime 0}v′=V′/V′0, we obtain the velocity of the particle in S ′ S ′ S^(')S^{\prime}S′ as
(3.161) v ′ = 1 1 − u v 2 ( v 1 1 − u 2 , v 2 − u , v 3 1 − u 2 ) (3.161) v ′ = 1 1 − u v 2 v 1 1 − u 2 , v 2 − u , v 3 1 − u 2 {:(3.161)v^(')=(1)/(1-uv^(2))(v^(1)sqrt(1-u^(2)),v^(2)-u,v^(3)sqrt(1-u^(2))):}\begin{equation*} \boldsymbol{v}^{\prime}=\frac{1}{1-u v^{2}}\left(v^{1} \sqrt{1-u^{2}}, v^{2}-u, v^{3} \sqrt{1-u^{2}}\right) \tag{3.161} \end{equation*}(3.161)v′=11−uv2(v11−u2,v2−u,v31−u2)
1.41
Let K K KKK and be the rest frame of the Earth and let K ′ K ′ K^(')K^{\prime}K′ be the momentary rest frame of the spaceship. Relative to K ′ K ′ K^(')K^{\prime}K′, the space ship has the velocity u ′ = 0 u ′ = 0 u^(')=0u^{\prime}=0u′=0 at the considered time, which, according to transformations of velocities and accelerations, i.e.,
u ′ = d x ′ d t ′ = u − v 1 − u v c 2 , a ′ = d u ′ d t ′ = d 2 x ′ d t ′ 2 = a γ ( v ) 3 ( 1 − u v c 2 ) 3 , γ ( v ) ≡ 1 1 − v 2 c 2 u ′ = d x ′ d t ′ = u − v 1 − u v c 2 , a ′ = d u ′ d t ′ = d 2 x ′ d t ′ 2 = a γ ( v ) 3 1 − u v c 2 3 , γ ( v ) ≡ 1 1 − v 2 c 2 u^(')=(dx^('))/(dt^('))=(u-v)/(1-(uv)/(c^(2))),quada^(')=(du^('))/(dt^('))=(d^(2)x^('))/(dt^('2))=(a)/(gamma(v)^(3)(1-(uv)/(c^(2)))^(3)),quad gamma(v)-=(1)/(sqrt(1-(v^(2))/(c^(2))))u^{\prime}=\frac{d x^{\prime}}{d t^{\prime}}=\frac{u-v}{1-\frac{u v}{c^{2}}}, \quad a^{\prime}=\frac{d u^{\prime}}{d t^{\prime}}=\frac{d^{2} x^{\prime}}{d t^{\prime 2}}=\frac{a}{\gamma(v)^{3}\left(1-\frac{u v}{c^{2}}\right)^{3}}, \quad \gamma(v) \equiv \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}u′=dx′dt′=u−v1−uvc2,a′=du′dt′=d2x′dt′2=aγ(v)3(1−uvc2)3,γ(v)≡11−v2c2,
where u = d x d t u = d x d t u=(dx)/(dt)u=\frac{d x}{d t}u=dxdt and a = d u d t = d 2 x d t 2 a = d u d t = d 2 x d t 2 a=(du)/(dt)=(d^(2)x)/(dt^(2))a=\frac{d u}{d t}=\frac{d^{2} x}{d t^{2}}a=dudt=d2xdt2, means that
(3.163) u = v and a = a ′ ( 1 − v 2 c 2 ) 3 / 2 (3.163) u = v  and  a = a ′ 1 − v 2 c 2 3 / 2 {:(3.163)u=v quad" and "quad a=a^(')(1-(v^(2))/(c^(2)))^(3//2):}\begin{equation*} u=v \quad \text { and } \quad a=a^{\prime}\left(1-\frac{v^{2}}{c^{2}}\right)^{3 / 2} \tag{3.163} \end{equation*}(3.163)u=v and a=a′(1−v2c2)3/2
Since a = d u d t a = d u d t a=(du)/(dt)a=\frac{d u}{d t}a=dudt and a ′ = g a ′ = g a^(')=ga^{\prime}=ga′=g, it follows that
(3.164) d u d t = g ( 1 − u 2 c 2 ) 3 / 2 (3.164) d u d t = g 1 − u 2 c 2 3 / 2 {:(3.164)(du)/(dt)=g(1-(u^(2))/(c^(2)))^(3//2):}\begin{equation*} \frac{d u}{d t}=g\left(1-\frac{u^{2}}{c^{2}}\right)^{3 / 2} \tag{3.164} \end{equation*}(3.164)dudt=g(1−u2c2)3/2
which with the initial condition u ( 0 ) = 0 u ( 0 ) = 0 u(0)=0u(0)=0u(0)=0 has the solution
(3.165) u = d x d t = g t 1 + ( g t c ) 2 (3.165) u = d x d t = g t 1 + g t c 2 {:(3.165)u=(dx)/(dt)=(gt)/(sqrt(1+((gt)/(c))^(2))):}\begin{equation*} u=\frac{d x}{d t}=\frac{g t}{\sqrt{1+\left(\frac{g t}{c}\right)^{2}}} \tag{3.165} \end{equation*}(3.165)u=dxdt=gt1+(gtc)2
a) Integration of Eq. (3.165) with the initial condition x ( 0 ) = 0 x ( 0 ) = 0 x(0)=0x(0)=0x(0)=0 gives
(3.166) x = c 2 g [ 1 + ( g t c ) 2 − 1 ] (3.166) x = c 2 g 1 + g t c 2 − 1 {:(3.166)x=(c^(2))/(g)[sqrt(1+((gt)/(c))^(2))-1]:}\begin{equation*} x=\frac{c^{2}}{g}\left[\sqrt{1+\left(\frac{g t}{c}\right)^{2}}-1\right] \tag{3.166} \end{equation*}(3.166)x=c2g[1+(gtc)2−1]
which is a hyperbola in the Minkowski diagram; the motion is said to be hyperbolic.
b) According to the clock hypothesis, it holds that d τ = d t ′ d τ = d t ′ d tau=dt^(')d \tau=d t^{\prime}dτ=dt′, where τ τ tau\tauτ is the proper time of the spaceship and t ′ t ′ t^(')t^{\prime}t′ is the time relative to K ′ K ′ K^(')K^{\prime}K′. This means that
(3.167) d τ d t = d t ′ d t = 1 γ ( u ) = { Eq. ( 3.165 ) } = 1 1 + ( g t c ) 2 (3.167) d τ d t = d t ′ d t = 1 γ ( u ) = {  Eq.  ( 3.165 ) } = 1 1 + g t c 2 {:(3.167)(d tau)/(dt)=(dt^('))/(dt)=(1)/(gamma(u))={" Eq. "(3.165)}=(1)/(sqrt(1+((gt)/(c))^(2))):}\begin{equation*} \frac{d \tau}{d t}=\frac{d t^{\prime}}{d t}=\frac{1}{\gamma(u)}=\{\text { Eq. }(3.165)\}=\frac{1}{\sqrt{1+\left(\frac{g t}{c}\right)^{2}}} \tag{3.167} \end{equation*}(3.167)dτdt=dt′dt=1γ(u)={ Eq. (3.165)}=11+(gtc)2
After the substitution g t c = sinh ⁡ ϕ g t c = sinh ⁡ ϕ (gt)/(c)=sinh phi\frac{g t}{c}=\sinh \phigtc=sinh⁡ϕ, Eq. (3.167) is easily integrated. If τ ( 0 ) = 0 τ ( 0 ) = 0 tau(0)=0\tau(0)=0τ(0)=0, then the solution is
(3.168) t = c g sinh ⁡ g τ c (3.168) t = c g sinh ⁡ g τ c {:(3.168)t=(c)/(g)sinh((g tau)/(c)):}\begin{equation*} t=\frac{c}{g} \sinh \frac{g \tau}{c} \tag{3.168} \end{equation*}(3.168)t=cgsinh⁡gτc
Inserting Eq. (3.168) into Eq. (3.166) gives
(3.169) x = c 2 g ( cosh ⁡ g τ c − 1 ) (3.169) x = c 2 g cosh ⁡ g τ c − 1 {:(3.169)x=(c^(2))/(g)(cosh((g tau)/(c))-1):}\begin{equation*} x=\frac{c^{2}}{g}\left(\cosh \frac{g \tau}{c}-1\right) \tag{3.169} \end{equation*}(3.169)x=c2g(cosh⁡gτc−1)
If we measure distances in light years and times in years, then c = 1 c = 1 c=1c=1c=1 light year/ year and g ≃ 1.05 g ≃ 1.05 g≃1.05g \simeq 1.05g≃1.05 light years / ( year ) 2 / (  year  ) 2 //(" year ")^(2)/(\text { year })^{2}/( year )2. For x A = 2500000 x A = 2500000 x_(A)=2500000x_{A}=2500000xA=2500000 light years, Eq. (3.169) gives the proper time
τ A = c g arcosh ( 1 + g x A c 2 ) ≃ c g ln ⁡ 2 g x A c 2 (3.170) ≈ 1 1.05 ⋅ ln ⁡ ( 2 ⋅ 1.05 ⋅ 2.5 ⋅ 10 6 ) years ≈ 14.7 years. τ A = c g arcosh 1 + g x A c 2 ≃ c g ln ⁡ 2 g x A c 2 (3.170) ≈ 1 1.05 ⋅ ln ⁡ 2 ⋅ 1.05 ⋅ 2.5 ⋅ 10 6  years  ≈ 14.7  years.  {:[tau_(A)=(c)/(g)arcosh(1+(gx_(A))/(c^(2)))≃(c)/(g)ln((2gx_(A))/(c^(2)))],[(3.170)~~(1)/(1.05)*ln(2*1.05*2.5*10^(6))" years "~~14.7" years. "]:}\begin{align*} \tau_{A}=\frac{c}{g} \operatorname{arcosh}\left(1+\frac{g x_{A}}{c^{2}}\right) & \simeq \frac{c}{g} \ln \frac{2 g x_{A}}{c^{2}} \\ & \approx \frac{1}{1.05} \cdot \ln \left(2 \cdot 1.05 \cdot 2.5 \cdot 10^{6}\right) \text { years } \approx 14.7 \text { years. } \tag{3.170} \end{align*}τA=cgarcosh(1+gxAc2)≃cgln⁡2gxAc2(3.170)≈11.05⋅ln⁡(2⋅1.05⋅2.5⋅106) years ≈14.7 years. 
Thus, the commander of the spaceship will be almost 55 years old when the spaceship reaches the Andromeda Galaxy.

1.42

With d m < 0 d m < 0 dm < 0d m<0dm<0 being the decrease in mass of the rocket and d m ′ d m ′ dm^(')d m^{\prime}dm′ the mass of the ejecta, the conservation of energy and momentum in the instantaneous rest frame of the rocket takes the form
m c 2 = ( m + d m ) γ ( d u ) c 2 + d m ′ γ ( − w ) c 2 (3.171) ≃ ( m + d m ) c 2 + d m ′ c 2 γ ( w ) 0 = ( m + d m ) γ ( d u ) d u − w d m ′ γ ( − w ) (3.172) ≃ m d u − w d m ′ γ ( w ) m c 2 = ( m + d m ) γ ( d u ) c 2 + d m ′ γ ( − w ) c 2 (3.171) ≃ ( m + d m ) c 2 + d m ′ c 2 γ ( w ) 0 = ( m + d m ) γ ( d u ) d u − w d m ′ γ ( − w ) (3.172) ≃ m d u − w d m ′ γ ( w ) {:[mc^(2)=(m+dm)gamma(du)c^(2)+dm^(')gamma(-w)c^(2)],[(3.171)≃(m+dm)c^(2)+dm^(')c^(2)gamma(w)],[0=(m+dm)gamma(du)du-wdm^(')gamma(-w)],[(3.172)≃mdu-wdm^(')gamma(w)]:}\begin{align*} m c^{2} & =(m+d m) \gamma(d u) c^{2}+d m^{\prime} \gamma(-w) c^{2} \\ & \simeq(m+d m) c^{2}+d m^{\prime} c^{2} \gamma(w) \tag{3.171}\\ 0 & =(m+d m) \gamma(d u) d u-w d m^{\prime} \gamma(-w) \\ & \simeq m d u-w d m^{\prime} \gamma(w) \tag{3.172} \end{align*}mc2=(m+dm)γ(du)c2+dm′γ(−w)c2(3.171)≃(m+dm)c2+dm′c2γ(w)0=(m+dm)γ(du)du−wdm′γ(−w)(3.172)≃mdu−wdm′γ(w)
where we have kept only terms to linear order in small quantities and d u d u dud udu is the change in velocity of the rocket in the instantaneous rest frame. Solving for d u d u dud udu in terms of d m d m dmd mdm, we find that
(3.173) d u = − w m d m (3.173) d u = − w m d m {:(3.173)du=-(w)/(m)dm:}\begin{equation*} d u=-\frac{w}{m} d m \tag{3.173} \end{equation*}(3.173)du=−wmdm
With the relative velocity of the instantaneous rest frame and K K KKK being v v vvv, we find that the change in velocity is given by relativistic addition of velocity
v + d v = v + d u 1 + v d u c 2 ≃ ( v + d u ) ( 1 − v d u c 2 ) ≃ v + d u ( 1 − v 2 c 2 ) (3.174) = v − w ( 1 − v 2 c 2 ) d m m v + d v = v + d u 1 + v d u c 2 ≃ ( v + d u ) 1 − v d u c 2 ≃ v + d u 1 − v 2 c 2 (3.174) = v − w 1 − v 2 c 2 d m m {:[v+dv=(v+du)/(1+(vdu)/(c^(2)))≃(v+du)(1-(vdu)/(c^(2)))≃v+du(1-(v^(2))/(c^(2)))],[(3.174)=v-w(1-(v^(2))/(c^(2)))(dm)/(m)]:}\begin{align*} v+d v & =\frac{v+d u}{1+\frac{v d u}{c^{2}}} \simeq(v+d u)\left(1-\frac{v d u}{c^{2}}\right) \simeq v+d u\left(1-\frac{v^{2}}{c^{2}}\right) \\ & =v-w\left(1-\frac{v^{2}}{c^{2}}\right) \frac{d m}{m} \tag{3.174} \end{align*}v+dv=v+du1+vduc2≃(v+du)(1−vduc2)≃v+du(1−v2c2)(3.174)=v−w(1−v2c2)dmm
where we again keep only linear terms in the small quantities. It follows that
(3.175) d m m = − d v w ( 1 − v 2 / c 2 ) (3.175) d m m = − d v w 1 − v 2 / c 2 {:(3.175)(dm)/(m)=-(dv)/(w(1-v^(2)//c^(2))):}\begin{equation*} \frac{d m}{m}=-\frac{d v}{w\left(1-v^{2} / c^{2}\right)} \tag{3.175} \end{equation*}(3.175)dmm=−dvw(1−v2/c2)
Integrating this equation with the condition m = m 0 m = m 0 m=m_(0)m=m_{0}m=m0 when v = 0 v = 0 v=0v=0v=0 leads to
(3.176) m ( v ) = m 0 ( c − v c + v ) c 2 w (3.176) m ( v ) = m 0 c − v c + v c 2 w {:(3.176)m(v)=m_(0)((c-v)/(c+v))^((c)/(2w)):}\begin{equation*} m(v)=m_{0}\left(\frac{c-v}{c+v}\right)^{\frac{c}{2 w}} \tag{3.176} \end{equation*}(3.176)m(v)=m0(c−vc+v)c2w
Figure 3.7 Space-station frame (a) and my frame (b).
1.43
In order to obtain the velocity that my friend has in my frame, which will give me the direction, use the formula for relativistic addition of velocities with u x = 0 u x = 0 u_(x)=0u_{x}=0ux=0 and u y = v u y = v u_(y)=vu_{y}=vuy=v (which are measured in the space-station frame), i.e.,
(3.177) u x ′ = u x + ( − v ) 1 + u x ( − v ) / c 2 = − v , u y ′ = u y γ ( 1 + u x ( − v ) / c 2 ) = v γ (3.177) u x ′ = u x + ( − v ) 1 + u x ( − v ) / c 2 = − v , u y ′ = u y γ 1 + u x ( − v ) / c 2 = v γ {:(3.177)u_(x)^(')=(u_(x)+(-v))/(1+u_(x)(-v)//c^(2))=-v","quadu_(y)^(')=(u_(y))/(gamma(1+u_(x)(-v)//c^(2)))=(v)/( gamma):}\begin{equation*} u_{x}^{\prime}=\frac{u_{x}+(-v)}{1+u_{x}(-v) / c^{2}}=-v, \quad u_{y}^{\prime}=\frac{u_{y}}{\gamma\left(1+u_{x}(-v) / c^{2}\right)}=\frac{v}{\gamma} \tag{3.177} \end{equation*}(3.177)ux′=ux+(−v)1+ux(−v)/c2=−v,uy′=uyγ(1+ux(−v)/c2)=vγ
where − v − v -v-v−v is also the velocity (along the x x xxx-axis) of the space station in my frame and γ = ( 1 − v 2 / c 2 ) − 1 / 2 γ = 1 − v 2 / c 2 − 1 / 2 gamma=(1-v^(2)//c^(2))^(-1//2)\gamma=\left(1-v^{2} / c^{2}\right)^{-1 / 2}γ=(1−v2/c2)−1/2. The angle θ θ theta\thetaθ is given by (see Figure 3.7)
(3.178) tan ⁡ θ = u x ′ u y ′ = − v v / γ = − γ (3.178) tan ⁡ θ = u x ′ u y ′ = − v v / γ = − γ {:(3.178)tan theta=(u_(x)^('))/(u_(y)^('))=(-v)/(v//gamma)=-gamma:}\begin{equation*} \tan \theta=\frac{u_{x}^{\prime}}{u_{y}^{\prime}}=\frac{-v}{v / \gamma}=-\gamma \tag{3.178} \end{equation*}(3.178)tan⁡θ=ux′uy′=−vv/γ=−γ
Thus, I should send the message to my friend in the direction
(3.179) θ = − arctan ⁡ γ = − arccot 1 − v 2 / c 2 (3.179) θ = − arctan ⁡ γ = − arccot 1 − v 2 / c 2 {:(3.179)theta=-arctan gamma=-arccotsqrt(1-v^(2)//c^(2)):}\begin{equation*} \theta=-\arctan \gamma=-\operatorname{arccot} \sqrt{1-v^{2} / c^{2}} \tag{3.179} \end{equation*}(3.179)θ=−arctan⁡γ=−arccot1−v2/c2
defined in my frame. Note that in the nonrelativistic limit (i.e., v ≪ c v ≪ c v≪cv \ll cv≪c ), the direction is θ nr = − π / 4 = − 45 ∘ θ nr = − π / 4 = − 45 ∘ theta_(nr)=-pi//4=-45^(@)\theta_{\mathrm{nr}}=-\pi / 4=-45^{\circ}θnr=−π/4=−45∘.

1.44

a) The 4 -velocity of the object is given by U = γ ( u ) ( c , u ) U = γ ( u ) ( c , u ) U=gamma(u)(c,u)U=\gamma(u)(c, \mathbf{u})U=γ(u)(c,u) while the 4 -velocity of the frame S ′ S ′ S^(')S^{\prime}S′ is given by V = γ ( v ) ( c , − v , 0 , 0 ) V = γ ( v ) ( c , − v , 0 , 0 ) V=gamma(v)(c,-v,0,0)V=\gamma(v)(c,-v, 0,0)V=γ(v)(c,−v,0,0).
b) In any inertial frame, the 4 -velocity of an object traveling at velocity u u u\mathbf{u}u is given by U = γ ( u ) ( c , u ) U = γ ( u ) ( c , u ) U=gamma(u)(c,u)U=\gamma(u)(c, \mathbf{u})U=γ(u)(c,u) as given in a). An object at rest in the inertial frame therefore has 4-velocity V = ( c , 0 ) V = ( c , 0 ) V=(c,0)V=(c, \boldsymbol{0})V=(c,0) and as a result
(3.180) U ⋅ V = c 2 γ ( u ) (3.180) U ⋅ V = c 2 γ ( u ) {:(3.180)U*V=c^(2)gamma(u):}\begin{equation*} U \cdot V=c^{2} \gamma(u) \tag{3.180} \end{equation*}(3.180)U⋅V=c2γ(u)
As this inner product is a Lorentz invariant, it may be computed in any frame.
c) Taking the inner product of U U UUU and V V VVV as defined in a) we obtain
U ⋅ V = c 2 γ ( u ′ ) = γ ( u ) γ ( v ) ( c 2 − v u 1 ) ⟹ γ ( u ′ ) = γ ( u ) γ ( v ) ( 1 − v u 1 c 2 ) U ⋅ V = c 2 γ u ′ = γ ( u ) γ ( v ) c 2 − v u 1 ⟹ γ u ′ = γ ( u ) γ ( v ) 1 − v u 1 c 2 U*V=c^(2)gamma(u^('))=gamma(u)gamma(v)(c^(2)-vu_(1))quad Longrightarrowquad gamma(u^('))=gamma(u)gamma(v)(1-(vu_(1))/(c^(2)))U \cdot V=c^{2} \gamma\left(u^{\prime}\right)=\gamma(u) \gamma(v)\left(c^{2}-v u_{1}\right) \quad \Longrightarrow \quad \gamma\left(u^{\prime}\right)=\gamma(u) \gamma(v)\left(1-\frac{v u_{1}}{c^{2}}\right)U⋅V=c2γ(u′)=γ(u)γ(v)(c2−vu1)⟹γ(u′)=γ(u)γ(v)(1−vu1c2).

1.45

From the definition of the 4 -velocity, we know that V 2 = 1 V 2 = 1 V^(2)=1V^{2}=1V2=1. Differentiating this relation with respect to the proper time τ τ tau\tauτ leads to
(3.182) 0 = d 1 d τ = d V 2 d τ = V ⋅ d V d τ = V ⋅ A (3.182) 0 = d 1 d τ = d V 2 d τ = V ⋅ d V d τ = V ⋅ A {:(3.182)0=(d1)/(d tau)=(dV^(2))/(d tau)=V*(dV)/(d tau)=V*A:}\begin{equation*} 0=\frac{d 1}{d \tau}=\frac{d V^{2}}{d \tau}=V \cdot \frac{d V}{d \tau}=V \cdot A \tag{3.182} \end{equation*}(3.182)0=d1dτ=dV2dτ=V⋅dVdτ=V⋅A
It follows that the 4 -velocity V V VVV and the 4-acceleration A A AAA are always perpendicular as V ⋅ A = 0 V ⋅ A = 0 V*A=0V \cdot A=0V⋅A=0.

1.46

a) With the distance to the space station being 1 light day in my rest frame and the station moving toward me with speed c / 4 c / 4 c//4c / 4c/4 in the same frame. The distance between the signal and the station will decrease at 5 c / 4 5 c / 4 5c//45 c / 45c/4. Hence, it will take
(3.183) 1 light day 5 c / 4 = 0.8 days, (3.183) 1  light day  5 c / 4 = 0.8  days,  {:(3.183)(1" light day ")/(5c//4)=0.8" days, ":}\begin{equation*} \frac{1 \text { light day }}{5 c / 4}=0.8 \text { days, } \tag{3.183} \end{equation*}(3.183)1 light day 5c/4=0.8 days, 
for the signal to reach the station in my rest frame. The station will then be at a distance of 0.8 light days away from me as this is the distance the signal traveled. The rescue ship then travels toward me with speed 3 c / 4 3 c / 4 3c//43 c / 43c/4 in the station's rest frame.
Using relativistic addition of velocities, the rescue ship's speed relative to me is
(3.184) v ′ = 3 c / 4 + c / 4 1 + 3 / 16 = 16 c 19 (3.184) v ′ = 3 c / 4 + c / 4 1 + 3 / 16 = 16 c 19 {:(3.184)v^(')=(3c//4+c//4)/(1+3//16)=(16 c)/(19):}\begin{equation*} v^{\prime}=\frac{3 c / 4+c / 4}{1+3 / 16}=\frac{16 c}{19} \tag{3.184} \end{equation*}(3.184)v′=3c/4+c/41+3/16=16c19
The time taken for the rescue ship to reach me is therefore
(3.185) 0.8 light days 16 c / 19 = 0.95 days. (3.185) 0.8  light days  16 c / 19 = 0.95  days.  {:(3.185)(0.8" light days ")/(16 c//19)=0.95" days. ":}\begin{equation*} \frac{0.8 \text { light days }}{16 c / 19}=0.95 \text { days. } \tag{3.185} \end{equation*}(3.185)0.8 light days 16c/19=0.95 days. 
In total, it therefore takes the rescue ship 0.8 + 0.95 = 1.75 0.8 + 0.95 = 1.75 0.8+0.95=1.750.8+0.95=1.750.8+0.95=1.75 days to reach me according to my clock.

1.47

Let us assume that the criminal passes the police officer at t = 0 t = 0 t=0t=0t=0 and describe the events from the inertial frame S S SSS. We use a coordinate system such that the worldline of the police ship is given by
(3.186) x p 2 − t 2 = 1 a 2 (3.186) x p 2 − t 2 = 1 a 2 {:(3.186)x_(p)^(2)-t^(2)=(1)/(a^(2)):}\begin{equation*} x_{p}^{2}-t^{2}=\frac{1}{a^{2}} \tag{3.186} \end{equation*}(3.186)xp2−t2=1a2
as this corresponds to a worldline with proper acceleration of a a aaa (this may be shown by studying the dependence of x p x p x_(p)x_{p}xp on t t ttt for small t t ttt ). As this worldline has x p ( t = 0 ) = 1 a x p ( t = 0 ) = 1 a x_(p)(t=0)=(1)/(a)x_{p}(t=0)=\frac{1}{a}xp(t=0)=1a, the worldline of the criminal must also fulfill x c ( t = 0 ) = 0 x c ( t = 0 ) = 0 x_(c)(t=0)=0x_{c}(t=0)=0xc(t=0)=0 and correspond to the worldline of an object with constant velocity v v vvv. Thus, the criminal's worldline is
(3.187) x c = v t + 1 a (3.187) x c = v t + 1 a {:(3.187)x_(c)=vt+(1)/(a):}\begin{equation*} x_{c}=v t+\frac{1}{a} \tag{3.187} \end{equation*}(3.187)xc=vt+1a
In order to deduce where the two worldlines intersect, we set x c = x p x c = x p x_(c)=x_(p)x_{c}=x_{p}xc=xp and solve for the time t t ttt at which this happens, we obtain
(3.188) t [ 2 v a − ( 1 − v 2 ) t ] = 0 ⇒ t 1 = 0 or t 2 = 2 v γ 2 a (3.188) t 2 v a − 1 − v 2 t = 0 ⇒ t 1 = 0  or  t 2 = 2 v γ 2 a {:(3.188)t[2(v)/(a)-(1-v^(2))t]=0quad=>quadt_(1)=0quad" or "quadt_(2)=2(vgamma^(2))/(a):}\begin{equation*} t\left[2 \frac{v}{a}-\left(1-v^{2}\right) t\right]=0 \quad \Rightarrow \quad t_{1}=0 \quad \text { or } \quad t_{2}=2 \frac{v \gamma^{2}}{a} \tag{3.188} \end{equation*}(3.188)t[2va−(1−v2)t]=0⇒t1=0 or t2=2vγ2a
The solution t 1 t 1 t_(1)t_{1}t1 corresponds to the pass where the pursuit starts and t 2 t 2 t_(2)t_{2}t2 to that where it ends.
a) The criminal is moving with velocity v v vvv relative to S S SSS and is therefore simply time dilated by a factor 1 / γ 1 / γ 1//gamma1 / \gamma1/γ. Thus, the pursuit takes
(3.189) Δ t c = t 2 − t 1 γ = 2 v γ a (3.189) Δ t c = t 2 − t 1 γ = 2 v γ a {:(3.189)Deltat_(c)=(t_(2)-t_(1))/(gamma)=2(v gamma)/(a):}\begin{equation*} \Delta t_{c}=\frac{t_{2}-t_{1}}{\gamma}=2 \frac{v \gamma}{a} \tag{3.189} \end{equation*}(3.189)Δtc=t2−t1γ=2vγa
according to the criminal.
b) We can parametrize the police officer's worldline using hyperbolic functions as
(3.190) x p = 1 a cosh ⁡ ( a τ ) , t = 1 a sinh ⁡ ( a τ ) , (3.190) x p = 1 a cosh ⁡ ( a τ ) , t = 1 a sinh ⁡ ( a τ ) , {:(3.190)x_(p)=(1)/(a)cosh(a tau)","quad t=(1)/(a)sinh(a tau)",":}\begin{equation*} x_{p}=\frac{1}{a} \cosh (a \tau), \quad t=\frac{1}{a} \sinh (a \tau), \tag{3.190} \end{equation*}(3.190)xp=1acosh⁡(aτ),t=1asinh⁡(aτ),
which fulfills x p 2 − t 2 = 1 / a 2 x p 2 − t 2 = 1 / a 2 x_(p)^(2)-t^(2)=1//a^(2)x_{p}^{2}-t^{2}=1 / a^{2}xp2−t2=1/a2. The line element d s d s dsd sds is then given by
(3.191) d s 2 = d t 2 − d x 2 = d τ 2 [ cosh 2 ⁡ ( a τ ) − sinh 2 ⁡ ( a τ ) ] = d τ 2 (3.191) d s 2 = d t 2 − d x 2 = d τ 2 cosh 2 ⁡ ( a τ ) − sinh 2 ⁡ ( a τ ) = d τ 2 {:(3.191)ds^(2)=dt^(2)-dx^(2)=dtau^(2)[cosh^(2)(a tau)-sinh^(2)(a tau)]=dtau^(2):}\begin{equation*} d s^{2}=d t^{2}-d x^{2}=d \tau^{2}\left[\cosh ^{2}(a \tau)-\sinh ^{2}(a \tau)\right]=d \tau^{2} \tag{3.191} \end{equation*}(3.191)ds2=dt2−dx2=dτ2[cosh2⁡(aτ)−sinh2⁡(aτ)]=dτ2
It follows that the parameter τ τ tau\tauτ can be taken as the proper time of the police officer. We obtain
(3.192) Δ t p = τ 2 − τ 1 = 1 a arsinh ( a t 2 ) − 1 a arsinh ( a t 1 ) = 1 a arsinh ( 2 v γ 2 ) (3.192) Δ t p = τ 2 − τ 1 = 1 a arsinh a t 2 − 1 a arsinh a t 1 = 1 a arsinh 2 v γ 2 {:(3.192)Deltat_(p)=tau_(2)-tau_(1)=(1)/(a)arsinh(at_(2))-(1)/(a)arsinh(at_(1))=(1)/(a)arsinh(2vgamma^(2)):}\begin{equation*} \Delta t_{p}=\tau_{2}-\tau_{1}=\frac{1}{a} \operatorname{arsinh}\left(a t_{2}\right)-\frac{1}{a} \operatorname{arsinh}\left(a t_{1}\right)=\frac{1}{a} \operatorname{arsinh}\left(2 v \gamma^{2}\right) \tag{3.192} \end{equation*}(3.192)Δtp=τ2−τ1=1aarsinh(at2)−1aarsinh(at1)=1aarsinh(2vγ2)
c) By time reversal symmetry in the criminal's rest frame, the relative velocity between the two at the end of the pursuit must be v v vvv (in the opposite direction to when the pursuit started).
For peace of mind, we note that for v → 0 v → 0 v rarr0v \rightarrow 0v→0, we find that
(3.193) Δ t c ≃ 2 v a ≃ 1 a arsinh ( 2 v γ 2 ) , (3.193) Δ t c ≃ 2 v a ≃ 1 a arsinh 2 v γ 2 , {:(3.193)Deltat_(c)≃2(v)/(a)≃(1)/(a)arsinh(2vgamma^(2))",":}\begin{equation*} \Delta t_{c} \simeq 2 \frac{v}{a} \simeq \frac{1}{a} \operatorname{arsinh}\left(2 v \gamma^{2}\right), \tag{3.193} \end{equation*}(3.193)Δtc≃2va≃1aarsinh(2vγ2),
as expected in the nonrelativistic limit.

1.48

a) The relations d t = d X sinh ⁡ ( a T ) + X a cosh ⁡ ( a T ) d T , d x = d X cosh ⁡ ( a T ) + d t = d X sinh ⁡ ( a T ) + X a cosh ⁡ ( a T ) d T , d x = d X cosh ⁡ ( a T ) + dt=dX sinh(aT)+Xa cosh(aT)dT,dx=dX cosh(aT)+d t=d X \sinh (a T)+X a \cosh (a T) d T, d x=d X \cosh (a T)+dt=dXsinh⁡(aT)+Xacosh⁡(aT)dT,dx=dXcosh⁡(aT)+ X a sinh ⁡ ( a T ) d T , d y = d Y , d z = d Z X a sinh ⁡ ( a T ) d T , d y = d Y , d z = d Z Xa sinh(aT)dT,dy=dY,dz=dZX a \sinh (a T) d T, d y=d Y, d z=d ZXasinh⁡(aT)dT,dy=dY,dz=dZ give
(3.194) d s 2 = d t 2 − d x 2 − d y 2 − d z 2 = ⋯ = ( X a ) 2 d T 2 − d X 2 − d Y 2 − d Z 2 (3.194) d s 2 = d t 2 − d x 2 − d y 2 − d z 2 = ⋯ = ( X a ) 2 d T 2 − d X 2 − d Y 2 − d Z 2 {:(3.194)ds^(2)=dt^(2)-dx^(2)-dy^(2)-dz^(2)=cdots=(Xa)^(2)dT^(2)-dX^(2)-dY^(2)-dZ^(2):}\begin{equation*} d s^{2}=d t^{2}-d x^{2}-d y^{2}-d z^{2}=\cdots=(X a)^{2} d T^{2}-d X^{2}-d Y^{2}-d Z^{2} \tag{3.194} \end{equation*}(3.194)ds2=dt2−dx2−dy2−dz2=⋯=(Xa)2dT2−dX2−dY2−dZ2
i.e., the nonzero components of the metric tensor are g T T = ( X a ) 2 g T T = ( X a ) 2 g_(TT)=(Xa)^(2)g_{T T}=(X a)^{2}gTT=(Xa)2 and g X X = g X X = g_(XX)=g_{X X}=gXX= g Y Y = g Z Z = − 1 g Y Y = g Z Z = − 1 g_(YY)=g_(ZZ)=-1g_{Y Y}=g_{Z Z}=-1gYY=gZZ=−1.
b) The components of this vector in the astronauts coordinate system are
(3.195) ( k ′ ) μ ′ = ∂ ( x ′ ) μ ′ ∂ x v k v (3.195) k ′ μ ′ = ∂ x ′ μ ′ ∂ x v k v {:(3.195)(k^('))^(mu^('))=(del(x^('))^(mu^(')))/(delx^(v))k^(v):}\begin{equation*} \left(k^{\prime}\right)^{\mu^{\prime}}=\frac{\partial\left(x^{\prime}\right)^{\mu^{\prime}}}{\partial x^{v}} k^{v} \tag{3.195} \end{equation*}(3.195)(k′)μ′=∂(x′)μ′∂xvkv
with ( x ′ ) μ ′ = ( T , X , Y , Z ) x ′ μ ′ = ( T , X , Y , Z ) (x^('))^(mu^('))=(T,X,Y,Z)\left(x^{\prime}\right)^{\mu^{\prime}}=(T, X, Y, Z)(x′)μ′=(T,X,Y,Z) and ( x v ) = ( t , x , y , z ) x v = ( t , x , y , z ) (x^(v))=(t,x,y,z)\left(x^{v}\right)=(t, x, y, z)(xv)=(t,x,y,z). We compute the matrix
(3.196) Λ = ( ∂ ( x ′ ) μ ′ ∂ x v ) = ( ∂ T ∂ t ⋯ ∂ T ∂ z ⋮ ⋱ ⋮ ∂ Z ∂ t ⋯ ∂ Z ∂ z ) (3.196) Λ = ∂ x ′ μ ′ ∂ x v = ∂ T ∂ t ⋯ ∂ T ∂ z ⋮ ⋱ ⋮ ∂ Z ∂ t ⋯ ∂ Z ∂ z {:(3.196)Lambda=((del(x^('))^(mu^(')))/(delx^(v)))=([(del T)/(del t),cdots,(del T)/(del z)],[vdots,ddots,vdots],[(del Z)/(del t),cdots,(del Z)/(del z)]):}\Lambda=\left(\frac{\partial\left(x^{\prime}\right)^{\mu^{\prime}}}{\partial x^{v}}\right)=\left(\begin{array}{ccc} \frac{\partial T}{\partial t} & \cdots & \frac{\partial T}{\partial z} \tag{3.196}\\ \vdots & \ddots & \vdots \\ \frac{\partial Z}{\partial t} & \cdots & \frac{\partial Z}{\partial z} \end{array}\right)(3.196)Λ=(∂(x′)μ′∂xv)=(∂T∂t⋯∂T∂z⋮⋱⋮∂Z∂t⋯∂Z∂z)
To find this we note that x 2 − t 2 = X 2 x 2 − t 2 = X 2 x^(2)-t^(2)=X^(2)x^{2}-t^{2}=X^{2}x2−t2=X2 and t / x = tanh ⁡ ( a T ) t / x = tanh ⁡ ( a T ) t//x=tanh(aT)t / x=\tanh (a T)t/x=tanh⁡(aT), i.e.,
(3.197) X = x 2 − t 2 , T = 1 a artanh ( t / x ) , Y = y , Z = z (3.197) X = x 2 − t 2 , T = 1 a artanh ( t / x ) , Y = y , Z = z {:(3.197)X=sqrt(x^(2)-t^(2))","quad T=(1)/(a)artanh(t//x)","quad Y=y","quad Z=z:}\begin{equation*} X=\sqrt{x^{2}-t^{2}}, \quad T=\frac{1}{a} \operatorname{artanh}(t / x), \quad Y=y, \quad Z=z \tag{3.197} \end{equation*}(3.197)X=x2−t2,T=1aartanh(t/x),Y=y,Z=z
Using this we find that
(3.198) Λ = ( x a ( x 2 − t 2 ) − t a ( x 2 − t 2 ) 0 0 − t x 2 − t 2 x x 2 − t 2 0 0 0 0 1 0 0 0 0 1 ) = ( cosh ⁡ ( a T ) a X − sinh ⁡ ( a T ) a X 0 0 − sinh ⁡ ( a T ) cosh ⁡ ( a T ) 0 0 0 0 1 0 0 0 0 1 ) (3.198) Λ = x a x 2 − t 2 − t a x 2 − t 2 0 0 − t x 2 − t 2 x x 2 − t 2 0 0 0 0 1 0 0 0 0 1 = cosh ⁡ ( a T ) a X − sinh ⁡ ( a T ) a X 0 0 − sinh ⁡ ( a T ) cosh ⁡ ( a T ) 0 0 0 0 1 0 0 0 0 1 {:(3.198)Lambda=([(x)/(a(x^(2)-t^(2))),-(t)/(a(x^(2)-t^(2))),0,0],[-(t)/(sqrt(x^(2)-t^(2))),(x)/(sqrt(x^(2)-t^(2))),0,0],[0,0,1,0],[0,0,0,1])=([(cosh(aT))/(aX),-(sinh(aT))/(aX),0,0],[-sinh(aT),cosh(aT),0,0],[0,0,1,0],[0,0,0,1]):}\Lambda=\left(\begin{array}{cccc} \frac{x}{a\left(x^{2}-t^{2}\right)} & -\frac{t}{a\left(x^{2}-t^{2}\right)} & 0 & 0 \tag{3.198}\\ -\frac{t}{\sqrt{x^{2}-t^{2}}} & \frac{x}{\sqrt{x^{2}-t^{2}}} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)=\left(\begin{array}{cccc} \frac{\cosh (a T)}{a X} & -\frac{\sinh (a T)}{a X} & 0 & 0 \\ -\sinh (a T) & \cosh (a T) & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)(3.198)Λ=(xa(x2−t2)−ta(x2−t2)00−tx2−t2xx2−t20000100001)=(cosh⁡(aT)aX−sinh⁡(aT)aX00−sinh⁡(aT)cosh⁡(aT)0000100001)
and thus, we obtain
(3.199) ( ( k ′ ) T ( k ′ ) X ( k ′ ) Y ( k ′ ) Z ) = Λ ( ω ω cos ⁡ ( θ ) 0 ω sin ⁡ ( θ ) ) = ω ( ( 1 / a X ) [ cosh ⁡ ( a T ) − sinh ⁡ ( a T ) cos ⁡ ( θ ) ] − sinh ⁡ ( a T ) + cosh ⁡ ( a T ) cos ⁡ ( θ ) 0 sin ⁡ ( θ ) ) (3.199) k ′ T k ′ X k ′ Y k ′ Z = Λ ω ω cos ⁡ ( θ ) 0 ω sin ⁡ ( θ ) = ω ( 1 / a X ) [ cosh ⁡ ( a T ) − sinh ⁡ ( a T ) cos ⁡ ( θ ) ] − sinh ⁡ ( a T ) + cosh ⁡ ( a T ) cos ⁡ ( θ ) 0 sin ⁡ ( θ ) {:(3.199)([(k^('))^(T)],[(k^('))^(X)],[(k^('))^(Y)],[(k^('))^(Z)])=Lambda([omega],[omega cos(theta)],[0],[omega sin(theta)])=omega([(1//aX)[cosh(aT)-sinh(aT)cos(theta)]],[-sinh(aT)+cosh(aT)cos(theta)],[0],[sin(theta)]):}\left(\begin{array}{c} \left(k^{\prime}\right)^{T} \tag{3.199}\\ \left(k^{\prime}\right)^{X} \\ \left(k^{\prime}\right)^{Y} \\ \left(k^{\prime}\right)^{Z} \end{array}\right)=\Lambda\left(\begin{array}{c} \omega \\ \omega \cos (\theta) \\ 0 \\ \omega \sin (\theta) \end{array}\right)=\omega\left(\begin{array}{c} (1 / a X)[\cosh (a T)-\sinh (a T) \cos (\theta)] \\ -\sinh (a T)+\cosh (a T) \cos (\theta) \\ 0 \\ \sin (\theta) \end{array}\right)(3.199)((k′)T(k′)X(k′)Y(k′)Z)=Λ(ωωcos⁡(θ)0ωsin⁡(θ))=ω((1/aX)[cosh⁡(aT)−sinh⁡(aT)cos⁡(θ)]−sinh⁡(aT)+cosh⁡(aT)cos⁡(θ)0sin⁡(θ))
c) Inserting t ( T ) = X 0 sinh ⁡ ( a T ) , x ( T ) = X 0 cosh ⁡ ( a T ) , y ( T ) = v T t ( T ) = X 0 sinh ⁡ ( a T ) , x ( T ) = X 0 cosh ⁡ ( a T ) , y ( T ) = v T t(T)=X_(0)sinh(aT),x(T)=X_(0)cosh(aT),y(T)=vTt(T)=X_{0} \sinh (a T), x(T)=X_{0} \cosh (a T), y(T)=v Tt(T)=X0sinh⁡(aT),x(T)=X0cosh⁡(aT),y(T)=vT, and z ( T ) = 0 z ( T ) = 0 z(T)=0z(T)=0z(T)=0, we obtain
(3.200) τ = ∫ 0 T 0 t ′ ( T ) 2 − x ′ ( T ) 2 − y ′ ( T ) 2 d T = T 0 ( X 0 a ) 2 − v 2 (3.200) τ = ∫ 0 T 0   t ′ ( T ) 2 − x ′ ( T ) 2 − y ′ ( T ) 2 d T = T 0 X 0 a 2 − v 2 {:(3.200)tau=int_(0)^(T_(0))sqrt(t^(')(T)^(2)-x^(')(T)^(2)-y^(')(T)^(2))dT=T_(0)sqrt((X_(0)a)^(2)-v^(2)):}\begin{equation*} \tau=\int_{0}^{T_{0}} \sqrt{t^{\prime}(T)^{2}-x^{\prime}(T)^{2}-y^{\prime}(T)^{2}} d T=T_{0} \sqrt{\left(X_{0} a\right)^{2}-v^{2}} \tag{3.200} \end{equation*}(3.200)τ=∫0T0t′(T)2−x′(T)2−y′(T)2dT=T0(X0a)2−v2
1.49
By Lorentz invariance of the quantity x 2 − t 2 = 1 / α 2 x 2 − t 2 = 1 / α 2 x^(2)-t^(2)=1//alpha^(2)x^{2}-t^{2}=1 / \alpha^{2}x2−t2=1/α2, it follows that x ′ 2 − t ′ 2 = x ′ 2 − t ′ 2 = x^('2)-t^('2)=x^{\prime 2}-t^{\prime 2}=x′2−t′2= 1 / α 2 1 / α 2 1//alpha^(2)1 / \alpha^{2}1/α2. Differentiating this once with respect to t ′ t ′ t^(')t^{\prime}t′ gives
(3.201) 2 x ′ d x ′ d t ′ − 2 t ′ = 0 ⟹ v ′ = t ′ x ′ (3.201) 2 x ′ d x ′ d t ′ − 2 t ′ = 0 ⟹ v ′ = t ′ x ′ {:(3.201)2x^(')(dx^('))/(dt^('))-2t^(')=0quad Longrightarrowquadv^(')=(t^('))/(x^(')):}\begin{equation*} 2 x^{\prime} \frac{d x^{\prime}}{d t^{\prime}}-2 t^{\prime}=0 \quad \Longrightarrow \quad v^{\prime}=\frac{t^{\prime}}{x^{\prime}} \tag{3.201} \end{equation*}(3.201)2x′dx′dt′−2t′=0⟹v′=t′x′
A second differentiation with respect to t ′ t ′ t^(')t^{\prime}t′ results in
(3.202) v ′ 2 + x ′ a ′ − 1 = 0 ⟹ a ′ = 1 − v ′ 2 x ′ (3.202) v ′ 2 + x ′ a ′ − 1 = 0 ⟹ a ′ = 1 − v ′ 2 x ′ {:(3.202)v^('2)+x^(')a^(')-1=0quad Longrightarrowquada^(')=(1-v^('2))/(x^(')):}\begin{equation*} v^{\prime 2}+x^{\prime} a^{\prime}-1=0 \quad \Longrightarrow \quad a^{\prime}=\frac{1-v^{\prime 2}}{x^{\prime}} \tag{3.202} \end{equation*}(3.202)v′2+x′a′−1=0⟹a′=1−v′2x′
For t ′ = 0 t ′ = 0 t^(')=0t^{\prime}=0t′=0, we find that x ′ = 1 / α x ′ = 1 / α x^(')=1//alphax^{\prime}=1 / \alphax′=1/α and v ′ = 0 v ′ = 0 v^(')=0v^{\prime}=0v′=0. It follows that
(3.203) a ′ = α (3.203) a ′ = α {:(3.203)a^(')=alpha:}\begin{equation*} a^{\prime}=\alpha \tag{3.203} \end{equation*}(3.203)a′=α
for t ′ = 0 t ′ = 0 t^(')=0t^{\prime}=0t′=0. Thus, the acceleration in S ′ S ′ S^(')S^{\prime}S′ at time t ′ = 0 t ′ = 0 t^(')=0t^{\prime}=0t′=0 is the proper acceleration regardless of the relative velocity between the frames S S SSS and S ′ S ′ S^(')S^{\prime}S′.

1.50

The worldline of observer B B BBB may be described as
(3.204) x 2 − t 2 = 1 α 2 ⟹ x ( τ ) = 1 α cosh ⁡ ( α τ ) , t ( τ ) = 1 α sinh ⁡ ( α τ ) (3.204) x 2 − t 2 = 1 α 2 ⟹ x ( τ ) = 1 α cosh ⁡ ( α τ ) , t ( τ ) = 1 α sinh ⁡ ( α τ ) {:(3.204)x^(2)-t^(2)=(1)/(alpha^(2))quad Longrightarrowquad x(tau)=(1)/(alpha)cosh(alpha tau)","quad t(tau)=(1)/(alpha)sinh(alpha tau):}\begin{equation*} x^{2}-t^{2}=\frac{1}{\alpha^{2}} \quad \Longrightarrow \quad x(\tau)=\frac{1}{\alpha} \cosh (\alpha \tau), \quad t(\tau)=\frac{1}{\alpha} \sinh (\alpha \tau) \tag{3.204} \end{equation*}(3.204)x2−t2=1α2⟹x(τ)=1αcosh⁡(ατ),t(τ)=1αsinh⁡(ατ)
where τ τ tau\tauτ is the proper time since the start of acceleration and the acceleration starts at time t = 0 t = 0 t=0t=0t=0. We may of course add an arbitrary constant to the x x xxx-coordinate, but this is irrelevant for our purposes as we can simply put observer A A AAA at x = 1 / α x = 1 / α x=1//alphax=1 / \alphax=1/α. With the light signal being sent from A A AAA at t = t 0 t = t 0 t=t_(0)t=t_{0}t=t0, the worldline of the signal is given by
(3.205) x = 1 α + t − t 0 (3.205) x = 1 α + t − t 0 {:(3.205)x=(1)/(alpha)+t-t_(0):}\begin{equation*} x=\frac{1}{\alpha}+t-t_{0} \tag{3.205} \end{equation*}(3.205)x=1α+t−t0
The signal is received by B B BBB at the event where these worldlines cross and so we find that
(3.206) ( 1 α + t − t 0 ) 2 − t 2 = 1 α 2 ⟹ t = t 0 2 ⋅ 2 − α t 0 1 − α t 0 (3.206) 1 α + t − t 0 2 − t 2 = 1 α 2 ⟹ t = t 0 2 ⋅ 2 − α t 0 1 − α t 0 {:(3.206)((1)/(alpha)+t-t_(0))^(2)-t^(2)=(1)/(alpha^(2))Longrightarrow t=(t_(0))/(2)*(2-alphat_(0))/(1-alphat_(0)):}\begin{equation*} \left(\frac{1}{\alpha}+t-t_{0}\right)^{2}-t^{2}=\frac{1}{\alpha^{2}} \Longrightarrow t=\frac{t_{0}}{2} \cdot \frac{2-\alpha t_{0}}{1-\alpha t_{0}} \tag{3.206} \end{equation*}(3.206)(1α+t−t0)2−t2=1α2⟹t=t02⋅2−αt01−αt0
Using the expression for t t ttt in terms of the proper time, this leads to
(3.207) τ = 1 α arsinh ( α t 0 2 2 − α t 0 1 − α t 0 ) (3.207) τ = 1 α arsinh α t 0 2 2 − α t 0 1 − α t 0 {:(3.207)tau=(1)/(alpha)arsinh((alphat_(0))/(2)(2-alphat_(0))/(1-alphat_(0))):}\begin{equation*} \tau=\frac{1}{\alpha} \operatorname{arsinh}\left(\frac{\alpha t_{0}}{2} \frac{2-\alpha t_{0}}{1-\alpha t_{0}}\right) \tag{3.207} \end{equation*}(3.207)τ=1αarsinh(αt022−αt01−αt0)
For the case where α t 0 ≪ 1 α t 0 ≪ 1 alphat_(0)≪1\alpha t_{0} \ll 1αt0≪1, we can find a good approximation for the proper time τ τ tau\tauτ by expanding in this parameter with the result
(3.208) τ ≃ t 0 ( 1 + α t 0 2 ) (3.208) τ ≃ t 0 1 + α t 0 2 {:(3.208)tau≃t_(0)(1+(alphat_(0))/(2)):}\begin{equation*} \tau \simeq t_{0}\left(1+\frac{\alpha t_{0}}{2}\right) \tag{3.208} \end{equation*}(3.208)τ≃t0(1+αt02)
This is no surprise to us as the physical interpretation of the condition is that observer B B BBB has not had enough time to accelerate to a significant velocity, thus the main contribution to the proper time is given by the time t 0 t 0 t_(0)t_{0}t0.
In the case α t 0 → 1 α t 0 → 1 alphat_(0)rarr1\alpha t_{0} \rightarrow 1αt0→1, the time t t ttt as well as the proper time τ τ tau\tauτ for the signal arriving at B B BBB diverge. This is also relatively straightforward to understand as the asymptote of x 2 − t 2 = 1 / α 2 x 2 − t 2 = 1 / α 2 x^(2)-t^(2)=1//alpha^(2)x^{2}-t^{2}=1 / \alpha^{2}x2−t2=1/α2 is the line x = t x = t x=tx=tx=t. Therefore, if α t 0 ≥ 1 α t 0 ≥ 1 alphat_(0) >= 1\alpha t_{0} \geq 1αt0≥1, then the signal will never reach B B BBB.
Note that the solutions obtained for α t 0 > 1 α t 0 > 1 alphat_(0) > 1\alpha t_{0}>1αt0>1 are unphysical. They correspond to the intersections of the straight line with the other branch of x 2 − t 2 = 1 / α 2 x 2 − t 2 = 1 / α 2 x^(2)-t^(2)=1//alpha^(2)x^{2}-t^{2}=1 / \alpha^{2}x2−t2=1/α2.

1.51

The particles are moving in a circle with constant angular velocity ω ω omega\omegaω. Thus, we can write down an expression for the worldline (in the lab frame) of the particles by using the time t t ttt in the lab frame as the parameter
(3.209) x μ = ( t , R cos ⁡ ( ω t ) , R sin ⁡ ( ω t ) ) μ , (3.209) x μ = ( t , R cos ⁡ ( ω t ) , R sin ⁡ ( ω t ) ) μ , {:(3.209)x^(mu)=(t","R cos(omega t)","R sin(omega t))^(mu)",":}\begin{equation*} x^{\mu}=(t, R \cos (\omega t), R \sin (\omega t))^{\mu}, \tag{3.209} \end{equation*}(3.209)xμ=(t,Rcos⁡(ωt),Rsin⁡(ωt))μ,
where we have assumed the accelerator to have radius R R RRR and suppressed the z z zzz-coordinate which is constant. The velocity of the particles in the lab frame is
(3.210) v = d x d t = R ω ( − sin ⁡ ( ω t ) , cos ⁡ ( ω t ) ) ⟹ v = | v | = R ω = const. (3.210) v = d x d t = R ω ( − sin ⁡ ( ω t ) , cos ⁡ ( ω t ) ) ⟹ v = | v | = R ω =  const.  {:(3.210)v=(dx)/(dt)=R omega(-sin(omega t)","cos(omega t))quad Longrightarrowquad v=|v|=R omega=" const. ":}\begin{equation*} \boldsymbol{v}=\frac{d \boldsymbol{x}}{d t}=R \omega(-\sin (\omega t), \cos (\omega t)) \quad \Longrightarrow \quad v=|\boldsymbol{v}|=R \omega=\text { const. } \tag{3.210} \end{equation*}(3.210)v=dxdt=Rω(−sin⁡(ωt),cos⁡(ωt))⟹v=|v|=Rω= const. 
We now use the differential relation between the time in the lab frame and the eigentime of the particles
(3.211) d t d τ = γ ( v ) = 1 1 − v 2 = 1 1 − R 2 ω 2 (3.211) d t d τ = γ ( v ) = 1 1 − v 2 = 1 1 − R 2 ω 2 {:(3.211)(dt)/(d tau)=gamma(v)=(1)/(sqrt(1-v^(2)))=(1)/(sqrt(1-R^(2)omega^(2))):}\begin{equation*} \frac{d t}{d \tau}=\gamma(v)=\frac{1}{\sqrt{1-v^{2}}}=\frac{1}{\sqrt{1-R^{2} \omega^{2}}} \tag{3.211} \end{equation*}(3.211)dtdτ=γ(v)=11−v2=11−R2ω2
to obtain an expression for the 4 -velocity
(3.212) V μ = d x μ d τ = d t d τ d x μ d t = γ ( 1 , − R ω sin ⁡ ( ω t ) , R ω cos ⁡ ( ω t ) ) μ . (3.212) V μ = d x μ d τ = d t d τ d x μ d t = γ ( 1 , − R ω sin ⁡ ( ω t ) , R ω cos ⁡ ( ω t ) ) μ . {:(3.212)V^(mu)=(dx^(mu))/(d tau)=(dt)/(d tau)(dx^(mu))/(dt)=gamma(1","-R omega sin(omega t)","R omega cos(omega t))^(mu).:}\begin{equation*} V^{\mu}=\frac{d x^{\mu}}{d \tau}=\frac{d t}{d \tau} \frac{d x^{\mu}}{d t}=\gamma(1,-R \omega \sin (\omega t), R \omega \cos (\omega t))^{\mu} . \tag{3.212} \end{equation*}(3.212)Vμ=dxμdτ=dtdτdxμdt=γ(1,−Rωsin⁡(ωt),Rωcos⁡(ωt))μ.
Repeating the procedure to obtain the 4 -acceleration, we get
(3.213) A μ = d V μ d τ = d t d τ d V μ d t = − γ 2 R ω 2 ( 0 , cos ⁡ ( ω t ) , sin ⁡ ( ω t ) ) μ (3.213) A μ = d V μ d τ = d t d τ d V μ d t = − γ 2 R ω 2 ( 0 , cos ⁡ ( ω t ) , sin ⁡ ( ω t ) ) μ {:(3.213)A^(mu)=(dV^(mu))/(d tau)=(dt)/(d tau)(dV^(mu))/(dt)=-gamma^(2)Romega^(2)(0","cos(omega t)","sin(omega t))^(mu):}\begin{equation*} A^{\mu}=\frac{d V^{\mu}}{d \tau}=\frac{d t}{d \tau} \frac{d V^{\mu}}{d t}=-\gamma^{2} R \omega^{2}(0, \cos (\omega t), \sin (\omega t))^{\mu} \tag{3.213} \end{equation*}(3.213)Aμ=dVμdτ=dtdτdVμdt=−γ2Rω2(0,cos⁡(ωt),sin⁡(ωt))μ
since γ γ gamma\gammaγ is a constant. Note that this may also be written in terms of the orbital velocity v v vvv as
(3.214) A μ = − γ 2 v 2 R ⋅ ( 0 , cos ⁡ ( v t / R ) , sin ⁡ ( v t / R ) ) μ (3.214) A μ = − γ 2 v 2 R ⋅ ( 0 , cos ⁡ ( v t / R ) , sin ⁡ ( v t / R ) ) μ {:(3.214)A^(mu)=-gamma^(2)(v^(2))/(R)*(0","cos(vt//R)","sin(vt//R))^(mu):}\begin{equation*} A^{\mu}=-\gamma^{2} \frac{v^{2}}{R} \cdot(0, \cos (v t / R), \sin (v t / R))^{\mu} \tag{3.214} \end{equation*}(3.214)Aμ=−γ2v2R⋅(0,cos⁡(vt/R),sin⁡(vt/R))μ
if using v v vvv as a parameter instead of ω ω omega\omegaω. The proper acceleration a a aaa is given by − a 2 = A 2 − a 2 = A 2 -a^(2)=A^(2)-a^{2}=A^{2}−a2=A2 and thus
(3.215) a = − A 2 = γ 2 R ω 2 (3.215) a = − A 2 = γ 2 R ω 2 {:(3.215)a=sqrt(-A^(2))=gamma^(2)Romega^(2):}\begin{equation*} a=\sqrt{-A^{2}}=\gamma^{2} R \omega^{2} \tag{3.215} \end{equation*}(3.215)a=−A2=γ2Rω2
The eigentime required for the particles to complete one orbit is simply given by the relation between the eigentime and the lab time
(3.216) τ 0 = T γ = 2 π γ ω , (3.216) τ 0 = T γ = 2 π γ ω , {:(3.216)tau_(0)=(T)/( gamma)=(2pi)/(gamma omega)",":}\begin{equation*} \tau_{0}=\frac{T}{\gamma}=\frac{2 \pi}{\gamma \omega}, \tag{3.216} \end{equation*}(3.216)τ0=Tγ=2πγω,
where T = 2 π / ω T = 2 π / ω T=2pi//omegaT=2 \pi / \omegaT=2π/ω is the time to complete the orbit as measured in the lab, by integrating the differential relation taking into account the fact that γ γ gamma\gammaγ is constant.

1.52

The definition of the 4-force is
(3.217) F = d P d τ (3.217) F = d P d τ {:(3.217)F=(dP)/(d tau):}\begin{equation*} F=\frac{d P}{d \tau} \tag{3.217} \end{equation*}(3.217)F=dPdτ
where P = M V P = M V P=MVP=M VP=MV is the 4 -momentum and τ τ tau\tauτ the proper time of the object. The internal energy can be found by squaring the 4-momentum P 2 = M 2 V 2 = M 2 P 2 = M 2 V 2 = M 2 P^(2)=M^(2)V^(2)=M^(2)P^{2}=M^{2} V^{2}=M^{2}P2=M2V2=M2, since the square of the 4 -velocity is equal to one. It follows that
(3.218) d P 2 d τ = 2 P ⋅ d P d τ = 2 M d M d τ (3.218) d P 2 d τ = 2 P ⋅ d P d τ = 2 M d M d τ {:(3.218)(dP^(2))/(d tau)=2P*(dP)/(d tau)=2M(dM)/(d tau):}\begin{equation*} \frac{d P^{2}}{d \tau}=2 P \cdot \frac{d P}{d \tau}=2 M \frac{d M}{d \tau} \tag{3.218} \end{equation*}(3.218)dP2dτ=2P⋅dPdτ=2MdMdτ
where d P / d τ = F d P / d τ = F dP//d tau=Fd P / d \tau=FdP/dτ=F is the 4-force and d M / d τ d M / d τ dM//d taud M / d \taudM/dτ is the sought derivative of the internal energy with respect to the proper time. Using P = M V P = M V P=MVP=M VP=MV, we can solve for this quantity as
(3.219) d M d τ = V ⋅ F = f V ⋅ U = f γ (3.219) d M d τ = V ⋅ F = f V ⋅ U = f γ {:(3.219)(dM)/(d tau)=V*F=fV*U=f gamma:}\begin{equation*} \frac{d M}{d \tau}=V \cdot F=f V \cdot U=f \gamma \tag{3.219} \end{equation*}(3.219)dMdτ=V⋅F=fV⋅U=fγ
where γ = V ⋅ U γ = V ⋅ U gamma=V*U\gamma=V \cdot Uγ=V⋅U is the gamma factor for the relative velocity v v vvv between objects moving with the 4 -velocities U U UUU and V V VVV. (Alternatively, we note that there is a frame where U = ( 1 , 0 ) U = ( 1 , 0 ) U=(1,0)U=(1,0)U=(1,0). In this frame, V = γ ( 1 , v ) V = γ ( 1 , v ) V=gamma(1,v)V=\gamma(1, \boldsymbol{v})V=γ(1,v), where v v v\boldsymbol{v}v is the velocity of the object in this frame.)
By using the product rule on the derivative F = d P / d τ F = d P / d τ F=dP//d tauF=d P / d \tauF=dP/dτ, we find that
(3.220) F = d M d τ V + M d V d τ = f γ V + M A ⟹ A = f M ( U − γ V ) (3.220) F = d M d τ V + M d V d τ = f γ V + M A ⟹ A = f M ( U − γ V ) {:(3.220)F=(dM)/(d tau)V+M(dV)/(d tau)=f gamma V+MA quad Longrightarrowquad A=(f)/(M)(U-gamma V):}\begin{equation*} F=\frac{d M}{d \tau} V+M \frac{d V}{d \tau}=f \gamma V+M A \quad \Longrightarrow \quad A=\frac{f}{M}(U-\gamma V) \tag{3.220} \end{equation*}(3.220)F=dMdτV+MdVdτ=fγV+MA⟹A=fM(U−γV)
where A = d V / d τ A = d V / d τ A=dV//d tauA=d V / d \tauA=dV/dτ is the 4-acceleration of the object. The relation between the 4-acceleration A A AAA and the proper acceleration α α alpha\alphaα is
α 2 = − A 2 = − f 2 M 2 ⋅ ( U − γ V ) 2 = − f 2 M 2 ( U 2 − 2 γ V ⋅ U + γ 2 V 2 ) (3.221) = − f 2 M 2 ( 1 − 2 γ 2 + γ 2 ) = f 2 M 2 ( γ 2 − 1 ) = f 2 M 2 γ 2 v 2 . α 2 = − A 2 = − f 2 M 2 ⋅ ( U − γ V ) 2 = − f 2 M 2 U 2 − 2 γ V ⋅ U + γ 2 V 2 (3.221) = − f 2 M 2 1 − 2 γ 2 + γ 2 = f 2 M 2 γ 2 − 1 = f 2 M 2 γ 2 v 2 . {:[alpha^(2)=-A^(2)=-(f^(2))/(M^(2))*(U-gamma V)^(2)=-(f^(2))/(M^(2))(U^(2)-2gamma V*U+gamma^(2)V^(2))],[(3.221)=-(f^(2))/(M^(2))(1-2gamma^(2)+gamma^(2))=(f^(2))/(M^(2))(gamma^(2)-1)=(f^(2))/(M^(2))gamma^(2)v^(2).]:}\begin{align*} \alpha^{2} & =-A^{2}=-\frac{f^{2}}{M^{2}} \cdot(U-\gamma V)^{2}=-\frac{f^{2}}{M^{2}}\left(U^{2}-2 \gamma V \cdot U+\gamma^{2} V^{2}\right) \\ & =-\frac{f^{2}}{M^{2}}\left(1-2 \gamma^{2}+\gamma^{2}\right)=\frac{f^{2}}{M^{2}}\left(\gamma^{2}-1\right)=\frac{f^{2}}{M^{2}} \gamma^{2} v^{2} . \tag{3.221} \end{align*}α2=−A2=−f2M2⋅(U−γV)2=−f2M2(U2−2γV⋅U+γ2V2)(3.221)=−f2M2(1−2γ2+γ2)=f2M2(γ2−1)=f2M2γ2v2.
Thus, the proper acceleration is
(3.222) α = f M γ v (3.222) α = f M γ v {:(3.222)alpha=(f)/(M)gamma v:}\begin{equation*} \alpha=\frac{f}{M} \gamma v \tag{3.222} \end{equation*}(3.222)α=fMγv
In particular, we note that α = 0 α = 0 alpha=0\alpha=0α=0 if v = 0 v = 0 v=0v=0v=0, which is stating that the object is not accelerating if the 4 -force is parallel to the 4 -velocity, as expected.
1.53
With the 4-force being the derivative of the 4-momentum with respect to the proper time, we find that
(3.223) F μ = d P μ d τ = d ( m V μ ) d τ = d m d τ V μ + m A μ (3.223) F μ = d P μ d τ = d m V μ d τ = d m d τ V μ + m A μ {:(3.223)F^(mu)=(dP^(mu))/(d tau)=(d(mV^(mu)))/(d tau)=(dm)/(d tau)V^(mu)+mA^(mu):}\begin{equation*} F^{\mu}=\frac{d P^{\mu}}{d \tau}=\frac{d\left(m V^{\mu}\right)}{d \tau}=\frac{d m}{d \tau} V^{\mu}+m A^{\mu} \tag{3.223} \end{equation*}(3.223)Fμ=dPμdτ=d(mVμ)dτ=dmdτVμ+mAμ
where V μ V μ V^(mu)V^{\mu}Vμ is the 4 -velocity and A μ A μ A^(mu)A^{\mu}Aμ the 4 -acceleration. We can find the rate of change in the rest energy by taking the inner product with the 4 -velocity, resulting in
(3.224) F ⋅ V = d m d τ V 2 + m A ⋅ V = d m d τ (3.224) F ⋅ V = d m d τ V 2 + m A ⋅ V = d m d τ {:(3.224)F*V=(dm)/(d tau)V^(2)+mA*V=(dm)/(d tau):}\begin{equation*} F \cdot V=\frac{d m}{d \tau} V^{2}+m A \cdot V=\frac{d m}{d \tau} \tag{3.224} \end{equation*}(3.224)F⋅V=dmdτV2+mA⋅V=dmdτ
since V 2 = 1 V 2 = 1 V^(2)=1V^{2}=1V2=1 and A ⋅ V = 0 A ⋅ V = 0 A*V=0A \cdot V=0A⋅V=0. Using the given expression for F μ F μ F^(mu)F^{\mu}Fμ and that V μ = γ ( 1 , v ) μ V μ = γ ( 1 , v ) μ V^(mu)=gamma(1,v)^(mu)V^{\mu}=\gamma(1, \boldsymbol{v})^{\mu}Vμ=γ(1,v)μ, we now obtain
(3.225) d m d τ = ( 0 , f ) ⋅ γ ( 1 , v ) = − γ f ⋅ v (3.225) d m d τ = ( 0 , f ) ⋅ γ ( 1 , v ) = − γ f ⋅ v {:(3.225)(dm)/(d tau)=(0","f)*gamma(1","v)=-gamma f*v:}\begin{equation*} \frac{d m}{d \tau}=(0, \boldsymbol{f}) \cdot \gamma(1, \boldsymbol{v})=-\gamma \boldsymbol{f} \cdot \boldsymbol{v} \tag{3.225} \end{equation*}(3.225)dmdτ=(0,f)⋅γ(1,v)=−γf⋅v
Squaring the 4 -force, we find the relation
(3.226) F 2 = ( d m d τ V + m A ) 2 = ( d m d τ ) 2 + m 2 A 2 = γ 2 ( f ⋅ v ) 2 − m 2 α 2 (3.226) F 2 = d m d τ V + m A 2 = d m d τ 2 + m 2 A 2 = γ 2 ( f ⋅ v ) 2 − m 2 α 2 {:(3.226)F^(2)=((dm)/(d tau)V+mA)^(2)=((dm)/(d tau))^(2)+m^(2)A^(2)=gamma^(2)(f*v)^(2)-m^(2)alpha^(2):}\begin{equation*} F^{2}=\left(\frac{d m}{d \tau} V+m A\right)^{2}=\left(\frac{d m}{d \tau}\right)^{2}+m^{2} A^{2}=\gamma^{2}(\boldsymbol{f} \cdot \boldsymbol{v})^{2}-m^{2} \alpha^{2} \tag{3.226} \end{equation*}(3.226)F2=(dmdτV+mA)2=(dmdτ)2+m2A2=γ2(f⋅v)2−m2α2
Obviously, we also have F 2 = − f 2 F 2 = − f 2 F^(2)=-f^(2)F^{2}=-f^{2}F2=−f2 and so we can solve for the proper acceleration α α alpha\alphaα as
(3.227) α 2 = 1 m 2 [ f 2 + γ 2 ( f ⋅ v ) 2 ] (3.227) α 2 = 1 m 2 f 2 + γ 2 ( f ⋅ v ) 2 {:(3.227)alpha^(2)=(1)/(m^(2))[f^(2)+gamma^(2)(f*v)^(2)]:}\begin{equation*} \alpha^{2}=\frac{1}{m^{2}}\left[f^{2}+\gamma^{2}(\boldsymbol{f} \cdot \boldsymbol{v})^{2}\right] \tag{3.227} \end{equation*}(3.227)α2=1m2[f2+γ2(f⋅v)2]
The force is a pure force only if the rate d m / d τ = 0 d m / d τ = 0 dm//d tau=0d m / d \tau=0dm/dτ=0, corresponding to f ⋅ v = 0 f ⋅ v = 0 f*v=0\boldsymbol{f} \cdot \boldsymbol{v}=0f⋅v=0. Therefore, the force is a pure force if f f fff and v v vvv are orthogonal.

1.54

The 4 -acceleration is generally given by
(3.228) A = d V d τ = γ [ d γ d t ( 1 , v ) + γ ( 0 , a ) ] = γ 4 v ⋅ a ( 1 , v ) + γ 2 ( 0 , a ) (3.228) A = d V d τ = γ d γ d t ( 1 , v ) + γ ( 0 , a ) = γ 4 v ⋅ a ( 1 , v ) + γ 2 ( 0 , a ) {:(3.228)A=(dV)/(d tau)=gamma[(d gamma)/(dt)(1,v)+gamma(0,a)]=gamma^(4)v*a(1","v)+gamma^(2)(0","a):}\begin{equation*} A=\frac{d V}{d \tau}=\gamma\left[\frac{d \gamma}{d t}(1, \boldsymbol{v})+\gamma(0, \boldsymbol{a})\right]=\gamma^{4} \boldsymbol{v} \cdot \boldsymbol{a}(1, \boldsymbol{v})+\gamma^{2}(0, \boldsymbol{a}) \tag{3.228} \end{equation*}(3.228)A=dVdτ=γ[dγdt(1,v)+γ(0,a)]=γ4v⋅a(1,v)+γ2(0,a)
In the instantaneous rest frame, the 4-acceleration is therefore given by
(3.229) A = ( 0 , a 0 ) = ( 0 , a 0 x , a 0 y ) (3.229) A = 0 , a 0 = 0 , a 0 x , a 0 y {:(3.229)A=(0,a_(0))=(0,a_(0x),a_(0y)):}\begin{equation*} A=\left(0, a_{0}\right)=\left(0, a_{0 x}, a_{0 y}\right) \tag{3.229} \end{equation*}(3.229)A=(0,a0)=(0,a0x,a0y)
where we have suppressed the z z zzz-direction, which will behave just as the y y yyy-direction. By Lorentz transforming this to S ′ S ′ S^(')S^{\prime}S′, we find that
(3.230) A ′ = ( − v γ a 0 x , γ a 0 x , a 0 y ) (3.230) A ′ = − v γ a 0 x , γ a 0 x , a 0 y {:(3.230)A^(')=(-v gammaa_(0x),gammaa_(0x),a_(0y)):}\begin{equation*} A^{\prime}=\left(-v \gamma a_{0 x}, \gamma a_{0 x}, a_{0 y}\right) \tag{3.230} \end{equation*}(3.230)A′=(−vγa0x,γa0x,a0y)
We now use that the 4 -velocity in S ′ S ′ S^(')S^{\prime}S′ is given by V = γ ( 1 , − v , 0 ) V = γ ( 1 , − v , 0 ) V=gamma(1,-v,0)V=\gamma(1,-v, 0)V=γ(1,−v,0) to identify this with the general expression for the 4 -acceleration
(3.231) A ′ = − v γ 4 a x ′ ( 1 , − v , 0 ) + γ 2 ( 0 , a x ′ , a y ′ ) (3.231) A ′ = − v γ 4 a x ′ ( 1 , − v , 0 ) + γ 2 0 , a x ′ , a y ′ {:(3.231)A^(')=-vgamma^(4)a_(x)^(')(1","-v","0)+gamma^(2)(0,a_(x)^('),a_(y)^(')):}\begin{equation*} A^{\prime}=-v \gamma^{4} a_{x}^{\prime}(1,-v, 0)+\gamma^{2}\left(0, a_{x}^{\prime}, a_{y}^{\prime}\right) \tag{3.231} \end{equation*}(3.231)A′=−vγ4ax′(1,−v,0)+γ2(0,ax′,ay′)
which gives us
(3.232) a x ′ = γ − 3 a 0 x , a y ′ = γ − 2 a 0 y (3.232) a x ′ = γ − 3 a 0 x , a y ′ = γ − 2 a 0 y {:(3.232)a_(x)^(')=gamma^(-3)a_(0x)","quada_(y)^(')=gamma^(-2)a_(0y):}\begin{equation*} a_{x}^{\prime}=\gamma^{-3} a_{0 x}, \quad a_{y}^{\prime}=\gamma^{-2} a_{0 y} \tag{3.232} \end{equation*}(3.232)ax′=γ−3a0x,ay′=γ−2a0y
For a fixed proper acceleration α α alpha\alphaα, the acceleration in the instantaneous rest frame is given by a 0 x = α cos ⁡ ( θ ) a 0 x = α cos ⁡ ( θ ) a_(0x)=alpha cos(theta)a_{0 x}=\alpha \cos (\theta)a0x=αcos⁡(θ) and a 0 y = α sin ⁡ ( θ ) a 0 y = α sin ⁡ ( θ ) a_(0y)=alpha sin(theta)a_{0 y}=\alpha \sin (\theta)a0y=αsin⁡(θ), where θ θ theta\thetaθ is the angle between the acceleration and the x x xxx-direction. This results in an acceleration
(3.233) a ′ 2 = a ′ 2 = a x ′ 2 + a y ′ 2 = γ − 4 α 2 [ γ − 2 cos 2 ⁡ ( θ ) + sin 2 ⁡ ( θ ) ] (3.233) a ′ 2 = a ′ 2 = a x ′ 2 + a y ′ 2 = γ − 4 α 2 γ − 2 cos 2 ⁡ ( θ ) + sin 2 ⁡ ( θ ) {:(3.233)a^('2)=a^('2)=a_(x)^('2)+a_(y)^('2)=gamma^(-4)alpha^(2)[gamma^(-2)cos^(2)(theta)+sin^(2)(theta)]:}\begin{equation*} a^{\prime 2}=a^{\prime 2}=a_{x}^{\prime 2}+a_{y}^{\prime 2}=\gamma^{-4} \alpha^{2}\left[\gamma^{-2} \cos ^{2}(\theta)+\sin ^{2}(\theta)\right] \tag{3.233} \end{equation*}(3.233)a′2=a′2=ax′2+ay′2=γ−4α2[γ−2cos2⁡(θ)+sin2⁡(θ)]
Since γ ≥ 1 γ ≥ 1 gamma >= 1\gamma \geq 1γ≥1, the expression in the parenthesis varies between one [when sin 2 ⁡ ( θ ) = 1 sin 2 ⁡ ( θ ) = 1 sin^(2)(theta)=1\sin ^{2}(\theta)=1sin2⁡(θ)=1 ] and γ − 2 γ − 2 gamma^(-2)\gamma^{-2}γ−2 [when cos 2 ⁡ ( θ ) = 1 ] cos 2 ⁡ ( θ ) = 1 {:cos^(2)(theta)=1]\left.\cos ^{2}(\theta)=1\right]cos2⁡(θ)=1]. Consequently, we find that
(3.234) a max ′ = α γ − 2 , a min ′ = α γ − 3 (3.234) a max ′ = α γ − 2 , a min ′ = α γ − 3 {:(3.234)a_(max)^(')=alphagamma^(-2)","quada_(min)^(')=alphagamma^(-3):}\begin{equation*} a_{\max }^{\prime}=\alpha \gamma^{-2}, \quad a_{\min }^{\prime}=\alpha \gamma^{-3} \tag{3.234} \end{equation*}(3.234)amax′=αγ−2,amin′=αγ−3

1.55

In the S ′ S ′ S^(')S^{\prime}S′ frame, we can extract the energy of the particle by taking the inner product of the 4 -momentum with the 4 -vector
(3.235) V ′ = ( 1 , 0 , 0 , 0 ) (3.235) V ′ = ( 1 , 0 , 0 , 0 ) {:(3.235)V^(')=(1","0","0","0):}\begin{equation*} V^{\prime}=(1,0,0,0) \tag{3.235} \end{equation*}(3.235)V′=(1,0,0,0)
which is the 4 -velocity of an object at rest in S ′ S ′ S^(')S^{\prime}S′. Similarly, the x ′ x ′ x^(')x^{\prime}x′-component of the 3 -momentum can be obtained by taking the inner product with the 4 -vector
(3.236) T ′ = ( 0 , − 1 , 0 , 0 ) (3.236) T ′ = ( 0 , − 1 , 0 , 0 ) {:(3.236)T^(')=(0","-1","0","0):}\begin{equation*} T^{\prime}=(0,-1,0,0) \tag{3.236} \end{equation*}(3.236)T′=(0,−1,0,0)
To see this, we note that the 4 -momentum in S ′ S ′ S^(')S^{\prime}S′ is given by P ′ = ( E ′ , p ′ ) P ′ = E ′ , p ′ P^(')=(E^('),p^('))P^{\prime}=\left(E^{\prime}, \boldsymbol{p}^{\prime}\right)P′=(E′,p′), resulting in
(3.237) V ′ ⋅ P ′ = ( 1 , 0 , 0 , 0 ) ⋅ ( E ′ , p ′ ) = E ′ (3.238) T ′ ⋅ P ′ = ( 0 , − 1 , 0 , 0 ) ⋅ ( E ′ , p ′ ) = − ( − 1 ) p x ′ = p x ′ (3.237) V ′ ⋅ P ′ = ( 1 , 0 , 0 , 0 ) ⋅ E ′ , p ′ = E ′ (3.238) T ′ ⋅ P ′ = ( 0 , − 1 , 0 , 0 ) ⋅ E ′ , p ′ = − ( − 1 ) p x ′ = p x ′ {:[(3.237)V^(')*P^(')=(1","0","0","0)*(E^('),p^('))=E^(')],[(3.238)T^(')*P^(')=(0","-1","0","0)*(E^('),p^('))=-(-1)p_(x)^(')=p_(x)^(')]:}\begin{align*} & V^{\prime} \cdot P^{\prime}=(1,0,0,0) \cdot\left(E^{\prime}, \boldsymbol{p}^{\prime}\right)=E^{\prime} \tag{3.237}\\ & T^{\prime} \cdot P^{\prime}=(0,-1,0,0) \cdot\left(E^{\prime}, \boldsymbol{p}^{\prime}\right)=-(-1) p_{x}^{\prime}=p_{x}^{\prime} \tag{3.238} \end{align*}(3.237)V′⋅P′=(1,0,0,0)⋅(E′,p′)=E′(3.238)T′⋅P′=(0,−1,0,0)⋅(E′,p′)=−(−1)px′=px′
Lorentz transforming V ′ V ′ V^(')V^{\prime}V′ and T ′ T ′ T^(')T^{\prime}T′ to the S S SSS frame, we find that
(3.239) V = γ ( 1 , v , 0 , 0 ) , T = − γ ( v , 1 , 0 , 0 ) (3.239) V = γ ( 1 , v , 0 , 0 ) , T = − γ ( v , 1 , 0 , 0 ) {:(3.239)V=gamma(1","v","0","0)","quad T=-gamma(v","1","0","0):}\begin{equation*} V=\gamma(1, v, 0,0), \quad T=-\gamma(v, 1,0,0) \tag{3.239} \end{equation*}(3.239)V=γ(1,v,0,0),T=−γ(v,1,0,0)
Since the inner products are Lorentz invariant, we can directly compute that
(3.240) E ′ = V ⋅ P = γ ( 1 , v , 0 , 0 ) ⋅ ( E , p ) = γ ( E − v p x ) (3.240) E ′ = V ⋅ P = γ ( 1 , v , 0 , 0 ) ⋅ ( E , p ) = γ E − v p x {:(3.240)E^(')=V*P=gamma(1","v","0","0)*(E","p)=gamma(E-vp_(x)):}\begin{equation*} E^{\prime}=V \cdot P=\gamma(1, v, 0,0) \cdot(E, \boldsymbol{p})=\gamma\left(E-v p_{x}\right) \tag{3.240} \end{equation*}(3.240)E′=V⋅P=γ(1,v,0,0)⋅(E,p)=γ(E−vpx)
In the same fashion, we also find that
(3.241) p x ′ = T ⋅ P = − γ ( v , 1 , 0 , 0 ) ⋅ ( E , p ) = γ ( p x − v E ) (3.241) p x ′ = T ⋅ P = − γ ( v , 1 , 0 , 0 ) ⋅ ( E , p ) = γ p x − v E {:(3.241)p_(x)^(')=T*P=-gamma(v","1","0","0)*(E","p)=gamma(p_(x)-vE):}\begin{equation*} p_{x}^{\prime}=T \cdot P=-\gamma(v, 1,0,0) \cdot(E, \boldsymbol{p})=\gamma\left(p_{x}-v E\right) \tag{3.241} \end{equation*}(3.241)px′=T⋅P=−γ(v,1,0,0)⋅(E,p)=γ(px−vE)
Furthermore, the velocity of a particle in an arbitrary frame is given by v = p / E v = p / E v=p//Ev=p / Ev=p/E, where p p p\boldsymbol{p}p is the 3-momentum and E E EEE the energy. In particular, in S ′ S ′ S^(')S^{\prime}S′, this results in the velocity component in the x ′ x ′ x^(')x^{\prime}x′-direction being
(3.242) v x ′ = p x ′ E ′ = p x − v E E − v p x (3.242) v x ′ = p x ′ E ′ = p x − v E E − v p x {:(3.242)v_(x)^(')=(p_(x)^('))/(E^('))=(p_(x)-vE)/(E-vp_(x)):}\begin{equation*} v_{x}^{\prime}=\frac{p_{x}^{\prime}}{E^{\prime}}=\frac{p_{x}-v E}{E-v p_{x}} \tag{3.242} \end{equation*}(3.242)vx′=px′E′=px−vEE−vpx

1.56

By definition of the 4-force, we know that
(3.243) F μ = d P μ d τ = f ( 1 , 1 ) μ (3.243) F μ = d P μ d τ = f ( 1 , 1 ) μ {:(3.243)F^(mu)=(dP^(mu))/(d tau)=f(1","1)^(mu):}\begin{equation*} F^{\mu}=\frac{d P^{\mu}}{d \tau}=f(1, \mathbf{1})^{\mu} \tag{3.243} \end{equation*}(3.243)Fμ=dPμdτ=f(1,1)μ
Integrating this relation with the initial condition P μ ( 0 ) = ( m 0 , 0 ) μ P μ ( 0 ) = m 0 , 0 μ P^(mu)(0)=(m_(0),0)^(mu)P^{\mu}(0)=\left(m_{0}, \mathbf{0}\right)^{\mu}Pμ(0)=(m0,0)μ leads to
(3.244) P μ ( τ ) = ( f τ + m 0 , f τ ) μ (3.244) P μ ( τ ) = f τ + m 0 , f τ μ {:(3.244)P^(mu)(tau)=(f tau+m_(0),f tau)^(mu):}\begin{equation*} P^{\mu}(\tau)=\left(f \tau+m_{0}, f \tau\right)^{\mu} \tag{3.244} \end{equation*}(3.244)Pμ(τ)=(fτ+m0,fτ)μ
The mass of the object at proper time τ τ tau\tauτ is therefore given by
(3.245) m ( τ ) = P ( τ ) 2 = ( f τ + m 0 ) 2 − f 2 τ 2 = m 0 ( 2 f τ + m 0 ) (3.245) m ( τ ) = P ( τ ) 2 = f τ + m 0 2 − f 2 τ 2 = m 0 2 f τ + m 0 {:(3.245)m(tau)=sqrt(P(tau)^(2))=sqrt((f tau+m_(0))^(2)-f^(2)tau^(2))=sqrt(m_(0)(2f tau+m_(0))):}\begin{equation*} m(\tau)=\sqrt{P(\tau)^{2}}=\sqrt{\left(f \tau+m_{0}\right)^{2}-f^{2} \tau^{2}}=\sqrt{m_{0}\left(2 f \tau+m_{0}\right)} \tag{3.245} \end{equation*}(3.245)m(τ)=P(τ)2=(fτ+m0)2−f2τ2=m0(2fτ+m0)
The relation between the coordinate time t t ttt and the proper time τ τ tau\tauτ is given by
(3.246) d t d τ = γ = E m = f τ + m 0 m 0 ( 2 f τ + m 0 ) (3.246) d t d τ = γ = E m = f τ + m 0 m 0 2 f τ + m 0 {:(3.246)(dt)/(d tau)=gamma=(E)/(m)=(f tau+m_(0))/(sqrt(m_(0)(2f tau+m_(0)))):}\begin{equation*} \frac{d t}{d \tau}=\gamma=\frac{E}{m}=\frac{f \tau+m_{0}}{\sqrt{m_{0}\left(2 f \tau+m_{0}\right)}} \tag{3.246} \end{equation*}(3.246)dtdτ=γ=Em=fτ+m0m0(2fτ+m0)
Integrating this differential equation with the initial condition t ( 0 ) = 0 t ( 0 ) = 0 t(0)=0t(0)=0t(0)=0 leads to
(3.247) t ( τ ) = ( τ 3 m 0 + 2 3 f ) m 0 ( 2 f τ + m 0 ) − 2 m 0 3 f (3.247) t ( τ ) = τ 3 m 0 + 2 3 f m 0 2 f τ + m 0 − 2 m 0 3 f {:(3.247)t(tau)=((tau)/(3m_(0))+(2)/(3f))sqrt(m_(0)(2f tau+m_(0)))-(2m_(0))/(3f):}\begin{equation*} t(\tau)=\left(\frac{\tau}{3 m_{0}}+\frac{2}{3 f}\right) \sqrt{m_{0}\left(2 f \tau+m_{0}\right)}-\frac{2 m_{0}}{3 f} \tag{3.247} \end{equation*}(3.247)t(τ)=(τ3m0+23f)m0(2fτ+m0)−2m03f
Note that the integral can be performed by making the ansatz that a primitive function of the integrand is given by
(3.248) g ( τ ) = ( A τ + B ) m 0 ( 2 f τ + m 0 ) (3.248) g ( τ ) = ( A τ + B ) m 0 2 f τ + m 0 {:(3.248)g(tau)=(A tau+B)sqrt(m_(0)(2f tau+m_(0))):}\begin{equation*} g(\tau)=(A \tau+B) \sqrt{m_{0}\left(2 f \tau+m_{0}\right)} \tag{3.248} \end{equation*}(3.248)g(τ)=(Aτ+B)m0(2fτ+m0)
and fixing A A AAA and B B BBB by comparing g ′ ( τ ) g ′ ( τ ) g^(')(tau)g^{\prime}(\tau)g′(τ) with the integrand.

1.57

The 4 -velocity of p p ppp is given by
(3.249) U μ = γ ( u ) ( 1 , − u cos ⁡ ( θ ) , − u sin ⁡ ( θ ) ) μ , (3.249) U μ = γ ( u ) ( 1 , − u cos ⁡ ( θ ) , − u sin ⁡ ( θ ) ) μ , {:(3.249)U^(mu)=gamma(u)(1","-u cos(theta)","-u sin(theta))^(mu)",":}\begin{equation*} U^{\mu}=\gamma(u)(1,-u \cos (\theta),-u \sin (\theta))^{\mu}, \tag{3.249} \end{equation*}(3.249)Uμ=γ(u)(1,−ucos⁡(θ),−usin⁡(θ))μ,
in S S SSS. Lorentz transforming this to the rest frame S ′ S ′ S^(')S^{\prime}S′ of the observer o o ooo, we find that
(3.250) U μ ′ = γ ( v ) γ ( u ) ( 1 + v u cos ⁡ ( θ ) , − u cos ⁡ ( θ ) − v , − u sin ⁡ ( θ ) γ ( v ) − 1 ) μ ′ (3.250) U μ ′ = γ ( v ) γ ( u ) 1 + v u cos ⁡ ( θ ) , − u cos ⁡ ( θ ) − v , − u sin ⁡ ( θ ) γ ( v ) − 1 μ ′ {:(3.250)U^(mu^('))=gamma(v)gamma(u)(1+vu cos(theta),-u cos(theta)-v,-u sin(theta)gamma(v)^(-1))^(mu^(')):}\begin{equation*} U^{\mu^{\prime}}=\gamma(v) \gamma(u)\left(1+v u \cos (\theta),-u \cos (\theta)-v,-u \sin (\theta) \gamma(v)^{-1}\right)^{\mu^{\prime}} \tag{3.250} \end{equation*}(3.250)Uμ′=γ(v)γ(u)(1+vucos⁡(θ),−ucos⁡(θ)−v,−usin⁡(θ)γ(v)−1)μ′
The speed u ′ u ′ u^(')u^{\prime}u′ of p p ppp in S ′ S ′ S^(')S^{\prime}S′ can be found by considering the time component of the 4-velocity U U UUU in S ′ S ′ S^(')S^{\prime}S′ as
(3.251) γ ( u ′ ) = 1 1 − u ′ 2 ⟹ u ′ = 1 − 1 γ ( u ′ ) 2 (3.251) γ u ′ = 1 1 − u ′ 2 ⟹ u ′ = 1 − 1 γ u ′ 2 {:(3.251)gamma(u^('))=(1)/(sqrt(1-u^('2)))quad Longrightarrowquadu^(')=sqrt(1-(1)/(gamma(u^('))^(2))):}\begin{equation*} \gamma\left(u^{\prime}\right)=\frac{1}{\sqrt{1-u^{\prime 2}}} \quad \Longrightarrow \quad u^{\prime}=\sqrt{1-\frac{1}{\gamma\left(u^{\prime}\right)^{2}}} \tag{3.251} \end{equation*}(3.251)γ(u′)=11−u′2⟹u′=1−1γ(u′)2
From the expression for U U UUU in S ′ S ′ S^(')S^{\prime}S′, we therefore have
γ ( u ′ ) 2 = [ 1 + v u cos ⁡ ( θ ) ] 2 ( 1 − v 2 ) ( 1 − u 2 ) ⟹ u ′ = [ 1 + v u cos ⁡ ( θ ) ] 2 − ( 1 − v 2 ) ( 1 − u 2 ) 1 + v u cos ⁡ ( θ ) γ u ′ 2 = [ 1 + v u cos ⁡ ( θ ) ] 2 1 − v 2 1 − u 2 ⟹ u ′ = [ 1 + v u cos ⁡ ( θ ) ] 2 − 1 − v 2 1 − u 2 1 + v u cos ⁡ ( θ ) gamma(u^('))^(2)=([1+vu cos(theta)]^(2))/((1-v^(2))(1-u^(2)))quad Longrightarrowquadu^(')=(sqrt([1+vu cos(theta)]^(2)-(1-v^(2))(1-u^(2))))/(1+vu cos(theta))\gamma\left(u^{\prime}\right)^{2}=\frac{[1+v u \cos (\theta)]^{2}}{\left(1-v^{2}\right)\left(1-u^{2}\right)} \quad \Longrightarrow \quad u^{\prime}=\frac{\sqrt{[1+v u \cos (\theta)]^{2}-\left(1-v^{2}\right)\left(1-u^{2}\right)}}{1+v u \cos (\theta)}γ(u′)2=[1+vucos⁡(θ)]2(1−v2)(1−u2)⟹u′=[1+vucos⁡(θ)]2−(1−v2)(1−u2)1+vucos⁡(θ).
The tangent of the angle θ ′ θ ′ theta^(')\theta^{\prime}θ′, can be expressed as the ratio between U 2 ′ U 2 ′ U^(2^('))U^{2^{\prime}}U2′ and U 1 ′ U 1 ′ U^(1^('))U^{1^{\prime}}U1′, i.e.,
(3.253) tan ⁡ ( θ ′ ) = U 2 ′ U 1 ′ = u sin ⁡ ( θ ) γ ( v ) [ u cos ⁡ ( θ ) + v ] (3.253) tan ⁡ θ ′ = U 2 ′ U 1 ′ = u sin ⁡ ( θ ) γ ( v ) [ u cos ⁡ ( θ ) + v ] {:(3.253)tan(theta^('))=(U^(2^(')))/(U^(1^(')))=(u sin(theta))/(gamma(v)[u cos(theta)+v]):}\begin{equation*} \tan \left(\theta^{\prime}\right)=\frac{U^{2^{\prime}}}{U^{1^{\prime}}}=\frac{u \sin (\theta)}{\gamma(v)[u \cos (\theta)+v]} \tag{3.253} \end{equation*}(3.253)tan⁡(θ′)=U2′U1′=usin⁡(θ)γ(v)[ucos⁡(θ)+v]
In the nonrelativistic limit u , v ≪ 1 u , v ≪ 1 u,v≪1u, v \ll 1u,v≪1, the above expressions become
(3.254) u ′ = v 2 + u 2 + 2 v u cos ⁡ ( θ ) and tan ⁡ ( θ ′ ) = u sin ⁡ ( θ ) u cos ⁡ ( θ ) + v (3.254) u ′ = v 2 + u 2 + 2 v u cos ⁡ ( θ )  and  tan ⁡ θ ′ = u sin ⁡ ( θ ) u cos ⁡ ( θ ) + v {:(3.254)u^(')=sqrt(v^(2)+u^(2)+2vu cos(theta))quad" and "quad tan(theta^('))=(u sin(theta))/(u cos(theta)+v):}\begin{equation*} u^{\prime}=\sqrt{v^{2}+u^{2}+2 v u \cos (\theta)} \quad \text { and } \quad \tan \left(\theta^{\prime}\right)=\frac{u \sin (\theta)}{u \cos (\theta)+v} \tag{3.254} \end{equation*}(3.254)u′=v2+u2+2vucos⁡(θ) and tan⁡(θ′)=usin⁡(θ)ucos⁡(θ)+v
to leading order. The relation for the speed u ′ u ′ u^(')u^{\prime}u′ is just the cosine theorem for classical addition of velocities and the expression for tan ⁡ ( θ ′ ) tan ⁡ θ ′ tan(theta^('))\tan \left(\theta^{\prime}\right)tan⁡(θ′) is the classical aberration formula.
In the limit u → 1 u → 1 u rarr1u \rightarrow 1u→1, we find that
(3.255) u ′ → 1 and tan ⁡ ( θ ′ ) → sin ⁡ ( θ ) cos ⁡ ( θ ) + v (3.255) u ′ → 1  and  tan ⁡ θ ′ → sin ⁡ ( θ ) cos ⁡ ( θ ) + v {:(3.255)u^(')rarr1quad" and "quad tan(theta^('))rarr(sin(theta))/(cos(theta)+v):}\begin{equation*} u^{\prime} \rightarrow 1 \quad \text { and } \quad \tan \left(\theta^{\prime}\right) \rightarrow \frac{\sin (\theta)}{\cos (\theta)+v} \tag{3.255} \end{equation*}(3.255)u′→1 and tan⁡(θ′)→sin⁡(θ)cos⁡(θ)+v
The first relation represents the invariance of the speed of light and the second is the relativistic aberration formula for light.

1.58

In S S SSS, the object's worldline is given by
(3.256) x ( t ) = a t 2 2 (3.256) x ( t ) = a t 2 2 {:(3.256)x(t)=(at^(2))/(2):}\begin{equation*} x(t)=\frac{a t^{2}}{2} \tag{3.256} \end{equation*}(3.256)x(t)=at22
and its velocity by v ( t ) = d x d t = a t v ( t ) = d x d t = a t v(t)=(dx)/(dt)=atv(t)=\frac{d x}{d t}=a tv(t)=dxdt=at. The speed v 0 v 0 v_(0)v_{0}v0 is obtained at time t 0 = v 0 a t 0 = v 0 a t_(0)=(v_(0))/(a)t_{0}=\frac{v_{0}}{a}t0=v0a. The proper time to reach this speed is therefore
(3.257) τ ( v 0 ) = ∫ 0 t 0 1 − v ( t ) 2 d t = ∫ 0 t 0 1 − a 2 t 2 d t (3.257) τ v 0 = ∫ 0 t 0   1 − v ( t ) 2 d t = ∫ 0 t 0   1 − a 2 t 2 d t {:(3.257)tau(v_(0))=int_(0)^(t_(0))sqrt(1-v(t)^(2))dt=int_(0)^(t_(0))sqrt(1-a^(2)t^(2))dt:}\begin{equation*} \tau\left(v_{0}\right)=\int_{0}^{t_{0}} \sqrt{1-v(t)^{2}} d t=\int_{0}^{t_{0}} \sqrt{1-a^{2} t^{2}} d t \tag{3.257} \end{equation*}(3.257)τ(v0)=∫0t01−v(t)2dt=∫0t01−a2t2dt
Substituting a t = sin ⁡ θ a t = sin ⁡ θ at=sin thetaa t=\sin \thetaat=sin⁡θ, we find that d t = 1 a cos ⁡ θ d θ d t = 1 a cos ⁡ θ d θ dt=(1)/(a)cos theta d thetad t=\frac{1}{a} \cos \theta d \thetadt=1acos⁡θdθ, and therefore,
τ ( v 0 ) = 1 a ⋅ ∫ 0 θ 0 cos 2 ⁡ θ d θ = 1 2 a ⋅ ∫ 0 θ 0 ( 1 + cos ⁡ 2 θ ) d θ = θ 0 2 a + 1 4 a ⋅ sin ⁡ 2 θ 0 (3.258) = 1 2 a ( θ 0 + sin ⁡ θ 0 cos ⁡ θ 0 ) τ v 0 = 1 a ⋅ ∫ 0 θ 0   cos 2 ⁡ θ d θ = 1 2 a ⋅ ∫ 0 θ 0   ( 1 + cos ⁡ 2 θ ) d θ = θ 0 2 a + 1 4 a ⋅ sin ⁡ 2 θ 0 (3.258) = 1 2 a θ 0 + sin ⁡ θ 0 cos ⁡ θ 0 {:[tau(v_(0))=(1)/(a)*int_(0)^(theta_(0))cos^(2)theta d theta=(1)/(2a)*int_(0)^(theta_(0))(1+cos 2theta)d theta=(theta_(0))/(2a)+(1)/(4a)*sin 2theta_(0)],[(3.258)=(1)/(2a)(theta_(0)+sin theta_(0)cos theta_(0))]:}\begin{align*} \tau\left(v_{0}\right)=\frac{1}{a} \cdot \int_{0}^{\theta_{0}} \cos ^{2} \theta d \theta & =\frac{1}{2 a} \cdot \int_{0}^{\theta_{0}}(1+\cos 2 \theta) d \theta=\frac{\theta_{0}}{2 a}+\frac{1}{4 a} \cdot \sin 2 \theta_{0} \\ & =\frac{1}{2 a}\left(\theta_{0}+\sin \theta_{0} \cos \theta_{0}\right) \tag{3.258} \end{align*}τ(v0)=1a⋅∫0θ0cos2⁡θdθ=12a⋅∫0θ0(1+cos⁡2θ)dθ=θ02a+14a⋅sin⁡2θ0(3.258)=12a(θ0+sin⁡θ0cos⁡θ0)
where θ 0 = arcsin ⁡ ( a t 0 ) = arcsin ⁡ v 0 θ 0 = arcsin ⁡ a t 0 = arcsin ⁡ v 0 theta_(0)=arcsin(at_(0))=arcsin v_(0)\theta_{0}=\arcsin \left(a t_{0}\right)=\arcsin v_{0}θ0=arcsin⁡(at0)=arcsin⁡v0. Inserting the expression for θ 0 θ 0 theta_(0)\theta_{0}θ0 leads to
(3.259) τ ( v 0 ) = 1 2 a ( arcsin ⁡ v 0 + v 0 1 − v 0 2 ) (3.259) τ v 0 = 1 2 a arcsin ⁡ v 0 + v 0 1 − v 0 2 {:(3.259)tau(v_(0))=(1)/(2a)(arcsin v_(0)+v_(0)sqrt(1-v_(0)^(2))):}\begin{equation*} \tau\left(v_{0}\right)=\frac{1}{2 a}\left(\arcsin v_{0}+v_{0} \sqrt{1-v_{0}^{2}}\right) \tag{3.259} \end{equation*}(3.259)τ(v0)=12a(arcsin⁡v0+v01−v02)
The proper acceleration of the object is given by the square of its 4 -acceleration. We find that
A = d V d τ = d 2 X d τ 2 = d t d τ d d t ( d t d τ d X d t ) (3.260) = ( d t d τ ) 2 d 2 X d t 2 + d X d t d t d τ ⋅ d d t ( d t d τ ) = γ 2 ( 0 , a ) + V d γ d t A = d V d τ = d 2 X d τ 2 = d t d τ d d t d t d τ d X d t (3.260) = d t d τ 2 d 2 X d t 2 + d X d t d t d τ ⋅ d d t d t d τ = γ 2 ( 0 , a ) + V d γ d t {:[A=(dV)/(d tau)=(d^(2)X)/(dtau^(2))=(dt)/(d tau)(d)/(dt)((dt)/(d tau)(dX)/(dt))],[(3.260)=((dt)/(d tau))^(2)(d^(2)X)/(dt^(2))+(dX)/(dt)(dt)/(d tau)*(d)/(dt)((dt)/(d tau))=gamma^(2)(0","a)+V(d gamma)/(dt)]:}\begin{align*} A=\frac{d V}{d \tau} & =\frac{d^{2} X}{d \tau^{2}}=\frac{d t}{d \tau} \frac{d}{d t}\left(\frac{d t}{d \tau} \frac{d X}{d t}\right) \\ & =\left(\frac{d t}{d \tau}\right)^{2} \frac{d^{2} X}{d t^{2}}+\frac{d X}{d t} \frac{d t}{d \tau} \cdot \frac{d}{d t}\left(\frac{d t}{d \tau}\right)=\gamma^{2}(0, a)+V \frac{d \gamma}{d t} \tag{3.260} \end{align*}A=dVdτ=d2Xdτ2=dtdτddt(dtdτdXdt)(3.260)=(dtdτ)2d2Xdt2+dXdtdtdτ⋅ddt(dtdτ)=γ2(0,a)+Vdγdt
This leads to A − V d γ d t = γ 2 ( 0 , a ) A − V d γ d t = γ 2 ( 0 , a ) A-V(d gamma)/(dt)=gamma^(2)(0,a)A-V \frac{d \gamma}{d t}=\gamma^{2}(0, a)A−Vdγdt=γ2(0,a) and squaring this expression, we find
(3.261) A 2 + V 2 ( d γ d t ) 2 = − α 2 + ( d γ d t ) 2 = − γ 4 a 2 (3.261) A 2 + V 2 d γ d t 2 = − α 2 + d γ d t 2 = − γ 4 a 2 {:(3.261)A^(2)+V^(2)((d gamma)/(dt))^(2)=-alpha^(2)+((d gamma)/(dt))^(2)=-gamma^(4)a^(2):}\begin{equation*} A^{2}+V^{2}\left(\frac{d \gamma}{d t}\right)^{2}=-\alpha^{2}+\left(\frac{d \gamma}{d t}\right)^{2}=-\gamma^{4} a^{2} \tag{3.261} \end{equation*}(3.261)A2+V2(dγdt)2=−α2+(dγdt)2=−γ4a2
where α α alpha\alphaα is the proper acceleration and we have used that A ⋅ V = 0 A ⋅ V = 0 A*V=0A \cdot V=0A⋅V=0. It follows that
(3.262) α 2 = γ 4 a 2 + ( d γ d t ) 2 (3.262) α 2 = γ 4 a 2 + d γ d t 2 {:(3.262)alpha^(2)=gamma^(4)a^(2)+((d gamma)/(dt))^(2):}\begin{equation*} \alpha^{2}=\gamma^{4} a^{2}+\left(\frac{d \gamma}{d t}\right)^{2} \tag{3.262} \end{equation*}(3.262)α2=γ4a2+(dγdt)2
Inserting the expression for γ γ gamma\gammaγ into the derivative leads to
(3.263) d γ d t = d d t 1 1 − a 2 t 2 = − 1 1 − a 2 t 2 a 2 t = − γ 3 v a (3.263) d γ d t = d d t 1 1 − a 2 t 2 = − 1 1 − a 2 t 2 a 2 t = − γ 3 v a {:(3.263)(d gamma)/(dt)=(d)/(dt)(1)/(sqrt(1-a^(2)t^(2)))=-(1)/(sqrt(1-a^(2)t^(2)))a^(2)t=-gamma^(3)va:}\begin{equation*} \frac{d \gamma}{d t}=\frac{d}{d t} \frac{1}{\sqrt{1-a^{2} t^{2}}}=-\frac{1}{\sqrt{1-a^{2} t^{2}}} a^{2} t=-\gamma^{3} v a \tag{3.263} \end{equation*}(3.263)dγdt=ddt11−a2t2=−11−a2t2a2t=−γ3va
We therefore find
α 2 = γ 4 a 2 + γ 6 v 2 a 2 = γ 4 a 2 ( 1 + γ 2 v 2 ) (3.264) = γ 4 a 2 1 − v 2 + v 2 1 − v 2 = γ 4 a 2 1 1 − v 2 = γ 6 a 2 α 2 = γ 4 a 2 + γ 6 v 2 a 2 = γ 4 a 2 1 + γ 2 v 2 (3.264) = γ 4 a 2 1 − v 2 + v 2 1 − v 2 = γ 4 a 2 1 1 − v 2 = γ 6 a 2 {:[alpha^(2)=gamma^(4)a^(2)+gamma^(6)v^(2)a^(2)=gamma^(4)a^(2)(1+gamma^(2)v^(2))],[(3.264)=gamma^(4)a^(2)(1-v^(2)+v^(2))/(1-v^(2))=gamma^(4)a^(2)(1)/(1-v^(2))=gamma^(6)a^(2)]:}\begin{align*} \alpha^{2} & =\gamma^{4} a^{2}+\gamma^{6} v^{2} a^{2}=\gamma^{4} a^{2}\left(1+\gamma^{2} v^{2}\right) \\ & =\gamma^{4} a^{2} \frac{1-v^{2}+v^{2}}{1-v^{2}}=\gamma^{4} a^{2} \frac{1}{1-v^{2}}=\gamma^{6} a^{2} \tag{3.264} \end{align*}α2=γ4a2+γ6v2a2=γ4a2(1+γ2v2)(3.264)=γ4a21−v2+v21−v2=γ4a211−v2=γ6a2
Taking the square root of this, we obtain
(3.265) α = γ 3 a = a 1 − a 2 t 2 3 . (3.265) α = γ 3 a = a 1 − a 2 t 2 3 . {:(3.265)alpha=gamma^(3)a=(a)/(sqrt(1-a^(2)t^(2)))^(3).:}\begin{equation*} \alpha=\gamma^{3} a=\frac{a}{\sqrt{1-a^{2} t^{2}}}{ }^{3} . \tag{3.265} \end{equation*}(3.265)α=γ3a=a1−a2t23.

1.59

The speed of a light signal relative to the water is given by u 0 = c / n u 0 = c / n u_(0)=c//nu_{0}=c / nu0=c/n, where n n nnn is the refractive index of water. Meanwhile, the speed of the water relative to the lab frame is v v vvv and application of the formula for relativistic addition of velocities then results in the speed of the light signal relative to the lab frame
(3.266) u = u 0 + v 1 + u 0 v c 2 ≃ u 0 + k v (3.266) u = u 0 + v 1 + u 0 v c 2 ≃ u 0 + k v {:(3.266)u=(u_(0)+v)/(1+(u_(0)v)/(c^(2)))≃u_(0)+kv:}\begin{equation*} u=\frac{u_{0}+v}{1+\frac{u_{0} v}{c^{2}}} \simeq u_{0}+k v \tag{3.266} \end{equation*}(3.266)u=u0+v1+u0vc2≃u0+kv
where k = 1 − u 0 2 c 2 = 1 − 1 n 2 k = 1 − u 0 2 c 2 = 1 − 1 n 2 k=1-(u_(0)^(2))/(c^(2))=1-(1)/(n^(2))k=1-\frac{u_{0}^{2}}{c^{2}}=1-\frac{1}{n^{2}}k=1−u02c2=1−1n2 and only linear terms in v ≪ c v ≪ c v≪cv \ll cv≪c have been kept, which is Fizeau's result.
1.60
See the solution to Problem 1.59, i.e., Fizeau's result. In the case of the water moving perpendicular to the light, we assume that the water is moving in the x x xxx-direction with speed v v vvv and the light is moving in the y y yyy-direction with speed c / n c / n c//nc / nc/n (when the water is standing still). In this case, the formula for addition of velocities reads
(3.267) u 2 = u ′ γ ( 1 + u 1 v / c 2 ) = { u 1 = 0 } = u ′ γ ≃ u ′ ( 1 + v 2 2 c 2 ) ≈ u ′ = c n (3.267) u 2 = u ′ γ 1 + u 1 v / c 2 = u 1 = 0 = u ′ γ ≃ u ′ 1 + v 2 2 c 2 ≈ u ′ = c n {:(3.267)u_(2)=(u^('))/(gamma(1+u_(1)v//c^(2)))={u_(1)=0}=(u^('))/(gamma)≃u^(')(1+(v^(2))/(2c^(2)))~~u^(')=(c)/(n):}\begin{equation*} u_{2}=\frac{u^{\prime}}{\gamma\left(1+u_{1} v / c^{2}\right)}=\left\{u_{1}=0\right\}=\frac{u^{\prime}}{\gamma} \simeq u^{\prime}\left(1+\frac{v^{2}}{2 c^{2}}\right) \approx u^{\prime}=\frac{c}{n} \tag{3.267} \end{equation*}(3.267)u2=u′γ(1+u1v/c2)={u1=0}=u′γ≃u′(1+v22c2)≈u′=cn
In other words, the correction enters only at second order in the water speed.

1.61

The redshift is maximal for when the source moves directly away from the observer. A lower bound for the speed of 3 C 9 is therefore determined as
(3.268) u min = x 2 − 1 x 2 + 1 c ≃ 0.80 c , where x = 3600 1215 (3.268) u min = x 2 − 1 x 2 + 1 c ≃ 0.80 c ,  where  x = 3600 1215 {:(3.268)u_(min)=(x^(2)-1)/(x^(2)+1)c≃0.80 c","quad" where "x=(3600)/(1215):}\begin{equation*} u_{\min }=\frac{x^{2}-1}{x^{2}+1} c \simeq 0.80 c, \quad \text { where } x=\frac{3600}{1215} \tag{3.268} \end{equation*}(3.268)umin=x2−1x2+1c≃0.80c, where x=36001215
is the ratio of the observed and emitted wavelengths, by solving for x x xxx from the relativistic Doppler shift formula.

1.62

The electromagnetic wave is E ( x ) = E 0 sin ⁡ 2 π ( x 1 λ − v t ) = sin ⁡ ( 2 π x 1 λ − 2 π ν t ) E ( x ) = E 0 sin ⁡ 2 π x 1 λ − v t = sin ⁡ 2 π x 1 λ − 2 π ν t E(x)=E_(0)sin 2pi((x^(1))/(lambda)-vt)=sin((2pix^(1))/(lambda)-2pi nu t)E(x)=E_{0} \sin 2 \pi\left(\frac{x^{1}}{\lambda}-v t\right)=\sin \left(\frac{2 \pi x^{1}}{\lambda}-2 \pi \nu t\right)E(x)=E0sin⁡2π(x1λ−vt)=sin⁡(2πx1λ−2πνt). The argument can be rewritten as follows
2 π x 1 λ − 2 π v t = − 2 π v t + 2 π x 1 λ = − ( 2 π v t − 2 π x 1 λ ) = − ( 2 π v t − 2 π v λ v x 1 ) = { ω = 2 π v and c = λ ν } = − ( ω t − ω c x 1 ) = − ( ω c c t − ω c x 1 ) = { x 0 = c t } = − ( ω c x 0 − ω c x 1 ) (3.269) = − ( ω c , ω c , 0 , 0 ) ⋅ ( x 0 , x 1 , x 2 , x 3 ) = − k μ x μ 2 π x 1 λ − 2 π v t = − 2 π v t + 2 π x 1 λ = − 2 π v t − 2 π x 1 λ = − 2 π v t − 2 π v λ v x 1 = { ω = 2 π v  and  c = λ ν } = − ω t − ω c x 1 = − ω c c t − ω c x 1 = x 0 = c t = − ω c x 0 − ω c x 1 (3.269) = − ω c , ω c , 0 , 0 ⋅ x 0 , x 1 , x 2 , x 3 = − k μ x μ {:[(2pix^(1))/(lambda)-2pi vt=-2pi vt+(2pix^(1))/(lambda)=-(2pi vt-(2pix^(1))/(lambda))=-(2pi vt-(2pi v)/(lambda v)x^(1))],[={omega=2pi v" and "c=lambda nu}=-(omega t-(omega )/(c)x^(1))=-((omega )/(c)ct-(omega )/(c)x^(1))],[={x^(0)=ct}=-((omega )/(c)x^(0)-(omega )/(c)x^(1))],[(3.269)=-((omega )/(c),(omega )/(c),0,0)*(x^(0),x^(1),x^(2),x^(3))=-k_(mu)x^(mu)]:}\begin{align*} \frac{2 \pi x^{1}}{\lambda}-2 \pi v t & =-2 \pi v t+\frac{2 \pi x^{1}}{\lambda}=-\left(2 \pi v t-\frac{2 \pi x^{1}}{\lambda}\right)=-\left(2 \pi v t-\frac{2 \pi v}{\lambda v} x^{1}\right) \\ & =\{\omega=2 \pi v \text { and } c=\lambda \nu\}=-\left(\omega t-\frac{\omega}{c} x^{1}\right)=-\left(\frac{\omega}{c} c t-\frac{\omega}{c} x^{1}\right) \\ & =\left\{x^{0}=c t\right\}=-\left(\frac{\omega}{c} x^{0}-\frac{\omega}{c} x^{1}\right) \\ & =-\left(\frac{\omega}{c}, \frac{\omega}{c}, 0,0\right) \cdot\left(x^{0}, x^{1}, x^{2}, x^{3}\right)=-k_{\mu} x^{\mu} \tag{3.269} \end{align*}2πx1λ−2πvt=−2πvt+2πx1λ=−(2πvt−2πx1λ)=−(2πvt−2πvλvx1)={ω=2πv and c=λν}=−(ωt−ωcx1)=−(ωcct−ωcx1)={x0=ct}=−(ωcx0−ωcx1)(3.269)=−(ωc,ωc,0,0)⋅(x0,x1,x2,x3)=−kμxμ
where k = ( ω c , ω c , 0 , 0 ) k = ω c , ω c , 0 , 0 k=((omega )/(c),(omega )/(c),0,0)k=\left(\frac{\omega}{c}, \frac{\omega}{c}, 0,0\right)k=(ωc,ωc,0,0) and x = ( x 0 , x 1 , x 2 , x 3 ) x = x 0 , x 1 , x 2 , x 3 x=(x^(0),x^(1),x^(2),x^(3))x=\left(x^{0}, x^{1}, x^{2}, x^{3}\right)x=(x0,x1,x2,x3). Thus, we can write E ( x ) = E ( x ) = E(x)=E(x)=E(x)= sin ⁡ ( − k μ x μ ) = − sin ⁡ k μ x μ sin ⁡ − k μ x μ = − sin ⁡ k μ x μ sin(-k_(mu)x^(mu))=-sin k_(mu)x^(mu)\sin \left(-k_{\mu} x^{\mu}\right)=-\sin k_{\mu} x^{\mu}sin⁡(−kμxμ)=−sin⁡kμxμ.
The wave vector k k kkk is lightlike, since k 2 = k μ k μ = ( ω c , ω c , 0 , 0 ) ⋅ ( ω c , ω c , 0 , 0 ) = k 2 = k μ k μ = ω c , ω c , 0 , 0 ⋅ ω c , ω c , 0 , 0 = k^(2)=k_(mu)k^(mu)=((omega )/(c),(omega )/(c),0,0)*((omega )/(c),(omega )/(c),0,0)=k^{2}=k_{\mu} k^{\mu}=\left(\frac{\omega}{c}, \frac{\omega}{c}, 0,0\right) \cdot\left(\frac{\omega}{c}, \frac{\omega}{c}, 0,0\right)=k2=kμkμ=(ωc,ωc,0,0)⋅(ωc,ωc,0,0)= ( ω c ) 2 − ( ω c ) 2 − 0 2 − 0 2 = 0 ω c 2 − ω c 2 − 0 2 − 0 2 = 0 ((omega )/(c))^(2)-((omega )/(c))^(2)-0^(2)-0^(2)=0\left(\frac{\omega}{c}\right)^{2}-\left(\frac{\omega}{c}\right)^{2}-0^{2}-0^{2}=0(ωc)2−(ωc)2−02−02=0.
In K ′ K ′ K^(')K^{\prime}K′, we have E ′ ( x ′ ) = − E 0 sin ⁡ k μ ′ x ′ μ = − E 0 sin ⁡ k μ x μ = E ( x ) E ′ x ′ = − E 0 sin ⁡ k μ ′ x ′ μ = − E 0 sin ⁡ k μ x μ = E ( x ) E^(')(x^('))=-E_(0)sin k_(mu)^(')x^('mu)=-E_(0)sin k_(mu)x^(mu)=E(x)E^{\prime}\left(x^{\prime}\right)=-E_{0} \sin k_{\mu}^{\prime} x^{\prime \mu}=-E_{0} \sin k_{\mu} x^{\mu}=E(x)E′(x′)=−E0sin⁡kμ′x′μ=−E0sin⁡kμxμ=E(x). Since this is Lorentz invariant, we find that k ′ = Λ k k ′ = Λ k k^(')=Lambda kk^{\prime}=\Lambda kk′=Λk, where Λ Λ Lambda\LambdaΛ is a Lorentz transformation, and
ω ′ c = k ′ 0 = k 0 cosh ⁡ θ − k 1 sinh ⁡ θ = ω c cosh ⁡ θ − ω c sinh ⁡ θ = ω c ( cosh ⁡ θ − sinh ⁡ θ ) ω ′ c = k ′ 0 = k 0 cosh ⁡ θ − k 1 sinh ⁡ θ = ω c cosh ⁡ θ − ω c sinh ⁡ θ = ω c ( cosh ⁡ θ − sinh ⁡ θ ) (omega^('))/(c)=k^('0)=k^(0)cosh theta-k^(1)sinh theta=(omega )/(c)cosh theta-(omega )/(c)sinh theta=(omega )/(c)(cosh theta-sinh theta)\frac{\omega^{\prime}}{c}=k^{\prime 0}=k^{0} \cosh \theta-k^{1} \sinh \theta=\frac{\omega}{c} \cosh \theta-\frac{\omega}{c} \sinh \theta=\frac{\omega}{c}(\cosh \theta-\sinh \theta)ω′c=k′0=k0cosh⁡θ−k1sinh⁡θ=ωccosh⁡θ−ωcsinh⁡θ=ωc(cosh⁡θ−sinh⁡θ).
Using the definitions of the hyperbolic functions and the fact that ω = 2 π ν ω = 2 π ν omega=2pi nu\omega=2 \pi \nuω=2πν and ω ′ = 2 π ν ′ ω ′ = 2 π ν ′ omega^(')=2pinu^(')\omega^{\prime}=2 \pi \nu^{\prime}ω′=2πν′, we obtain the answer
(3.271) v ′ = v e − θ (3.271) v ′ = v e − θ {:(3.271)v^(')=ve^(-theta):}\begin{equation*} v^{\prime}=v e^{-\theta} \tag{3.271} \end{equation*}(3.271)v′=ve−θ
which is the formula for the Doppler shift. If we instead use the relations cosh ⁡ θ = cosh ⁡ θ = cosh theta=\cosh \theta=cosh⁡θ= γ ( v ) γ ( v ) gamma(v)\gamma(v)γ(v) and sinh ⁡ θ = v c γ ( v ) sinh ⁡ θ = v c γ ( v ) sinh theta=(v)/(c)gamma(v)\sinh \theta=\frac{v}{c} \gamma(v)sinh⁡θ=vcγ(v), where γ ( v ) ≡ 1 1 − ( v c ) 2 γ ( v ) ≡ 1 1 − v c 2 gamma(v)-=(1)/(sqrt(1-((v)/(c))^(2)))\gamma(v) \equiv \frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^{2}}}γ(v)≡11−(vc)2, we instead obtain
(3.272) v ′ = v [ γ ( v ) − v c γ ( v ) ] = v γ ( v ) ( 1 − v c ) = v c − v c + v (3.272) v ′ = v γ ( v ) − v c γ ( v ) = v γ ( v ) 1 − v c = v c − v c + v {:(3.272)v^(')=v[gamma(v)-(v)/(c)gamma(v)]=v gamma(v)(1-(v)/(c))=vsqrt((c-v)/(c+v)):}\begin{equation*} v^{\prime}=v\left[\gamma(v)-\frac{v}{c} \gamma(v)\right]=v \gamma(v)\left(1-\frac{v}{c}\right)=v \sqrt{\frac{c-v}{c+v}} \tag{3.272} \end{equation*}(3.272)v′=v[γ(v)−vcγ(v)]=vγ(v)(1−vc)=vc−vc+v
which is the usual formula for the Doppler shift.

1.63

From the statement that the GRB has twice the duration as a nearby GRB, we conclude that the GRB is time dilated due to its motion and that γ = 2 γ = 2 gamma=2\gamma=2γ=2. From this relation, we can solve for the velocity of the GRB and obtain
(3.273) v = 3 2 (3.273) v = 3 2 {:(3.273)v=(sqrt3)/(2):}\begin{equation*} v=\frac{\sqrt{3}}{2} \tag{3.273} \end{equation*}(3.273)v=32
The quotient between the observed and emitted wavelengths is then given by the formula for the Doppler shift, i.e.,
(3.274) λ λ 0 = 1 + v 1 − v = 2 + 3 (3.274) λ λ 0 = 1 + v 1 − v = 2 + 3 {:(3.274)(lambda)/(lambda_(0))=sqrt((1+v)/(1-v))=2+sqrt3:}\begin{equation*} \frac{\lambda}{\lambda_{0}}=\sqrt{\frac{1+v}{1-v}}=2+\sqrt{3} \tag{3.274} \end{equation*}(3.274)λλ0=1+v1−v=2+3
Computing the redshift z z zzz, we obtain
(3.275) z ≡ λ − λ 0 λ 0 = 1 + 3 (3.275) z ≡ λ − λ 0 λ 0 = 1 + 3 {:(3.275)z-=(lambda-lambda_(0))/(lambda_(0))=1+sqrt3:}\begin{equation*} z \equiv \frac{\lambda-\lambda_{0}}{\lambda_{0}}=1+\sqrt{3} \tag{3.275} \end{equation*}(3.275)z≡λ−λ0λ0=1+3

1.64

The trajectories of objects 1 and 2 in the person's frame of reference K ( t , x , y , z ) K ( t , x , y , z ) K(t,x,y,z)K(t, x, y, z)K(t,x,y,z) are
(3.276) x 1 ( t ) = c t 2 , x 2 ( t ) = L − c t 2 , (3.276) x 1 ( t ) = c t 2 , x 2 ( t ) = L − c t 2 , {:(3.276)x_(1)(t)=(ct)/(2)","quadx_(2)(t)=L-(ct)/(2)",":}\begin{equation*} x_{1}(t)=\frac{c t}{2}, \quad x_{2}(t)=L-\frac{c t}{2}, \tag{3.276} \end{equation*}(3.276)x1(t)=ct2,x2(t)=L−ct2,
where we assume, without loss of generality, that the objects move parallel to the x x xxx-axis and that the first object is at the origin at time t = 0 ; y j ( t ) = z j ( t ) = 0 t = 0 ; y j ( t ) = z j ( t ) = 0 t=0;y_(j)(t)=z_(j)(t)=0t=0 ; y_{j}(t)=z_{j}(t)=0t=0;yj(t)=zj(t)=0, for j = 1 , 2 j = 1 , 2 j=1,2j=1,2j=1,2, and L > 0 L > 0 L > 0L>0L>0 is the distance between the objects at time t = 0 t = 0 t=0t=0t=0, of course. The frame of reference K ′ K ′ K^(')K^{\prime}K′ of the observer on the first object is related to K K KKK by a Lorentz transformation
(3.277) t ′ = γ ( t − v x / c 2 ) , x ′ = γ ( x − v t ) , γ = 1 1 − v 2 / c 2 (3.277) t ′ = γ t − v x / c 2 , x ′ = γ ( x − v t ) , γ = 1 1 − v 2 / c 2 {:(3.277)t^(')=gamma(t-vx//c^(2))","quadx^(')=gamma(x-vt)","quad gamma=1sqrt(1-v^(2)//c^(2)):}\begin{equation*} t^{\prime}=\gamma\left(t-v x / c^{2}\right), \quad x^{\prime}=\gamma(x-v t), \quad \gamma=1 \sqrt{1-v^{2} / c^{2}} \tag{3.277} \end{equation*}(3.277)t′=γ(t−vx/c2),x′=γ(x−vt),γ=11−v2/c2
and y ′ = y , z ′ = z y ′ = y , z ′ = z y^(')=y,z^(')=zy^{\prime}=y, z^{\prime}=zy′=y,z′=z with v = c / 2 v = c / 2 v=c//2v=c / 2v=c/2 so that x 1 ′ ( t ) = 0 x 1 ′ ( t ) = 0 x_(1)^(')(t)=0x_{1}^{\prime}(t)=0x1′(t)=0.
a) We parametrize the trajectory of the second object in K K KKK as
(3.278) t ( s ) = s , x ( s ) = L − c s 2 (3.278) t ( s ) = s , x ( s ) = L − c s 2 {:(3.278)t(s)=s","quad x(s)=L-(cs)/(2):}\begin{equation*} t(s)=s, \quad x(s)=L-\frac{c s}{2} \tag{3.278} \end{equation*}(3.278)t(s)=s,x(s)=L−cs2
and Lorentz transform this to K ′ K ′ K^(')K^{\prime}K′ such that
t ′ ( s ) = γ [ t ( s ) − ( c / 2 ) x ( s ) / c 2 ] , x ′ ( s ) = γ [ x ( s ) − ( c / 2 ) t ( s ) ] (3.279) γ = 1 1 − ( 1 / 2 ) 2 = 2 3 t ′ ( s ) = γ t ( s ) − ( c / 2 ) x ( s ) / c 2 , x ′ ( s ) = γ [ x ( s ) − ( c / 2 ) t ( s ) ] (3.279) γ = 1 1 − ( 1 / 2 ) 2 = 2 3 {:[t^(')(s)=gamma[t(s)-(c//2)x(s)//c^(2)]","quadx^(')(s)=gamma[x(s)-(c//2)t(s)]],[(3.279)gamma=(1)/(sqrt(1-(1//2)^(2)))=(2)/(sqrt3)]:}\begin{gather*} t^{\prime}(s)=\gamma\left[t(s)-(c / 2) x(s) / c^{2}\right], \quad x^{\prime}(s)=\gamma[x(s)-(c / 2) t(s)] \\ \gamma=\frac{1}{\sqrt{1-(1 / 2)^{2}}}=\frac{2}{\sqrt{3}} \tag{3.279} \end{gather*}t′(s)=γ[t(s)−(c/2)x(s)/c2],x′(s)=γ[x(s)−(c/2)t(s)](3.279)γ=11−(1/2)2=23
i.e.,
(3.280) t ′ ( s ) = 2 3 [ 5 s 4 − L 2 c ] , x ′ ( s ) = 2 3 ( L − c s ) (3.280) t ′ ( s ) = 2 3 5 s 4 − L 2 c , x ′ ( s ) = 2 3 ( L − c s ) {:(3.280)t^(')(s)=(2)/(sqrt3)[(5s)/(4)-(L)/(2c)]","quadx^(')(s)=(2)/(sqrt3)(L-cs):}\begin{equation*} t^{\prime}(s)=\frac{2}{\sqrt{3}}\left[\frac{5 s}{4}-\frac{L}{2 c}\right], \quad x^{\prime}(s)=\frac{2}{\sqrt{3}}(L-c s) \tag{3.280} \end{equation*}(3.280)t′(s)=23[5s4−L2c],x′(s)=23(L−cs)
We now can compute the velocity of the second object in K ′ K ′ K^(')K^{\prime}K′ such that
(3.281) d x ′ d t ′ = d x ′ / d s d t ′ / d s = − c 5 / 4 = − 4 c 5 (3.281) d x ′ d t ′ = d x ′ / d s d t ′ / d s = − c 5 / 4 = − 4 c 5 {:(3.281)(dx^('))/(dt^('))=(dx^(')//ds)/(dt^(')//ds)=(-c)/(5//4)=-(4c)/(5):}\begin{equation*} \frac{d x^{\prime}}{d t^{\prime}}=\frac{d x^{\prime} / d s}{d t^{\prime} / d s}=\frac{-c}{5 / 4}=-\frac{4 c}{5} \tag{3.281} \end{equation*}(3.281)dx′dt′=dx′/dsdt′/ds=−c5/4=−4c5
Note that a faster way to obtain this answer is to use relativistic addition of velocities
(3.282) v = v 1 + v 2 1 + v 1 v 2 / c 2 (3.282) v = v 1 + v 2 1 + v 1 v 2 / c 2 {:(3.282)v=(v_(1)+v_(2))/(1+v_(1)v_(2)//c^(2)):}\begin{equation*} v=\frac{v_{1}+v_{2}}{1+v_{1} v_{2} / c^{2}} \tag{3.282} \end{equation*}(3.282)v=v1+v21+v1v2/c2
with v 1 = v 2 = − c / 2 v 1 = v 2 = − c / 2 v_(1)=v_(2)=-c//2v_{1}=v_{2}=-c / 2v1=v2=−c/2.
b) The 4 -wavevector of the photon in the first object's rest frame is
(3.283) ( k ′ μ ) = ( k ′ 0 , k ′ 1 , 0 , 0 ) (3.283) k ′ μ = k ′ 0 , k ′ 1 , 0 , 0 {:(3.283)(k^('mu))=(k^('0),k^('1),0,0):}\begin{equation*} \left(k^{\prime \mu}\right)=\left(k^{\prime 0}, k^{\prime 1}, 0,0\right) \tag{3.283} \end{equation*}(3.283)(k′μ)=(k′0,k′1,0,0)
where k ′ 0 = k ′ 1 = 2 π / λ 0 k ′ 0 = k ′ 1 = 2 π / λ 0 k^('0)=k^('1)=2pi//lambda_(0)k^{\prime 0}=k^{\prime 1}=2 \pi / \lambda_{0}k′0=k′1=2π/λ0. The corresponding 4-wavevector in the second object's rest frame is
(3.284) ( k ′ ′ μ ) = ( k ′ ′ 0 , k ′ ′ 1 , 0 , 0 ) (3.284) k ′ ′ μ = k ′ ′ 0 , k ′ ′ 1 , 0 , 0 {:(3.284)(k^(''mu))=(k^(''0),k^(''1),0,0):}\begin{equation*} \left(k^{\prime \prime \mu}\right)=\left(k^{\prime \prime 0}, k^{\prime \prime 1}, 0,0\right) \tag{3.284} \end{equation*}(3.284)(k′′μ)=(k′′0,k′′1,0,0)
where
(3.285) k ′ ′ 0 = γ ~ ( k ′ 0 + v ~ k 1 / c ) , k ′ ′ 1 = γ ~ ( k ′ 1 + v ~ k ′ 0 / c ) , γ ~ = 1 / 1 − ( v ~ / c ) 2 (3.285) k ′ ′ 0 = γ ~ k ′ 0 + v ~ k 1 / c , k ′ ′ 1 = γ ~ k ′ 1 + v ~ k ′ 0 / c , γ ~ = 1 / 1 − ( v ~ / c ) 2 {:(3.285)k^(''0)= tilde(gamma)(k^('0)+( tilde(v))k^(1)//c)","quadk^(''1)= tilde(gamma)(k^('1)+( tilde(v))k^('0)//c)","quad tilde(gamma)=1//sqrt(1-( tilde(v)//c)^(2)):}\begin{equation*} k^{\prime \prime 0}=\tilde{\gamma}\left(k^{\prime 0}+\tilde{v} k^{1} / c\right), \quad k^{\prime \prime 1}=\tilde{\gamma}\left(k^{\prime 1}+\tilde{v} k^{\prime 0} / c\right), \quad \tilde{\gamma}=1 / \sqrt{1-(\tilde{v} / c)^{2}} \tag{3.285} \end{equation*}(3.285)k′′0=γ~(k′0+v~k1/c),k′′1=γ~(k′1+v~k′0/c),γ~=1/1−(v~/c)2
with v ~ = 4 c / 5 v ~ = 4 c / 5 tilde(v)=4c//5\tilde{v}=4 c / 5v~=4c/5, i.e.,
(3.286) ( k ′ ′ μ ) = ( k , k , 0 , 0 ) , k = 2 π λ 0 γ ~ ( 1 + v ~ / c ) = 2 π λ 0 1 + v ~ / c 1 − v ~ / c = 2 π λ 0 ⋅ 3 = 2 π λ (3.286) k ′ ′ μ = ( k , k , 0 , 0 ) , k = 2 π λ 0 γ ~ ( 1 + v ~ / c ) = 2 π λ 0 1 + v ~ / c 1 − v ~ / c = 2 π λ 0 ⋅ 3 = 2 π λ {:(3.286)(k^(''mu))=(k","k","0","0)","quad k=(2pi)/(lambda_(0)) tilde(gamma)(1+ tilde(v)//c)=(2pi)/(lambda_(0))sqrt((1+( tilde(v))//c)/(1-( tilde(v))//c))=(2pi)/(lambda_(0))*3=(2pi)/(lambda):}\begin{equation*} \left(k^{\prime \prime \mu}\right)=(k, k, 0,0), \quad k=\frac{2 \pi}{\lambda_{0}} \tilde{\gamma}(1+\tilde{v} / c)=\frac{2 \pi}{\lambda_{0}} \sqrt{\frac{1+\tilde{v} / c}{1-\tilde{v} / c}}=\frac{2 \pi}{\lambda_{0}} \cdot 3=\frac{2 \pi}{\lambda} \tag{3.286} \end{equation*}(3.286)(k′′μ)=(k,k,0,0),k=2πλ0γ~(1+v~/c)=2πλ01+v~/c1−v~/c=2πλ0⋅3=2πλ
Thus, the observer on the second object assigns the wavelength
(3.287) λ = λ 0 1 − v ~ / c 1 + v ~ / c = λ 0 3 ≃ 231.4 nm (3.287) λ = λ 0 1 − v ~ / c 1 + v ~ / c = λ 0 3 ≃ 231.4 nm {:(3.287)lambda=lambda_(0)sqrt((1-( tilde(v))//c)/(1+( tilde(v))//c))=(lambda_(0))/(3)≃231.4nm:}\begin{equation*} \lambda=\lambda_{0} \sqrt{\frac{1-\tilde{v} / c}{1+\tilde{v} / c}}=\frac{\lambda_{0}}{3} \simeq 231.4 \mathrm{~nm} \tag{3.287} \end{equation*}(3.287)λ=λ01−v~/c1+v~/c=λ03≃231.4 nm
to the photon. The wavelength is shorter: it has been shifted from red to UV, i.e., the frequency of the photon is higher.

1.65

a) In the rest frame of the observer, we can choose coordinates such that the light source is on the positive x x xxx-axis. The 4 -velocity of the light source can then be written as V = γ ( v ) ( 1 , v cos ⁡ ( θ ) , v sin ⁡ ( θ ) , 0 ) V = γ ( v ) ( 1 , v cos ⁡ ( θ ) , v sin ⁡ ( θ ) , 0 ) V=gamma(v)(1,v cos(theta),v sin(theta),0)V=\gamma(v)(1, v \cos (\theta), v \sin (\theta), 0)V=γ(v)(1,vcos⁡(θ),vsin⁡(θ),0) and the 4-frequency of the light pulse is given by N = ω ( 1 , − 1 , 0 , 0 ) N = ω ( 1 , − 1 , 0 , 0 ) N=omega(1,-1,0,0)N=\omega(1,-1,0,0)N=ω(1,−1,0,0). The frequency in the rest frame of the source is then obtained as
(3.288) ω 0 = V ⋅ N = γ ( v ) ω [ 1 + v cos ⁡ ( θ ) ] (3.288) ω 0 = V ⋅ N = γ ( v ) ω [ 1 + v cos ⁡ ( θ ) ] {:(3.288)omega_(0)=V*N=gamma(v)omega[1+v cos(theta)]:}\begin{equation*} \omega_{0}=V \cdot N=\gamma(v) \omega[1+v \cos (\theta)] \tag{3.288} \end{equation*}(3.288)ω0=V⋅N=γ(v)ω[1+vcos⁡(θ)]
Consequently, we find that
(3.289) ω = ω 0 1 − v 2 1 + v cos ⁡ ( θ ) (3.289) ω = ω 0 1 − v 2 1 + v cos ⁡ ( θ ) {:(3.289)omega=omega_(0)(sqrt(1-v^(2)))/(1+v cos(theta)):}\begin{equation*} \omega=\omega_{0} \frac{\sqrt{1-v^{2}}}{1+v \cos (\theta)} \tag{3.289} \end{equation*}(3.289)ω=ω01−v21+vcos⁡(θ)
In particular, when the light source is moving directly away from the observer, we have cos ⁡ ( θ ) = 1 cos ⁡ ( θ ) = 1 cos(theta)=1\cos (\theta)=1cos⁡(θ)=1 and therefore recover the relativistic Doppler formula
(3.290) ω = ω 0 ( 1 − v ) ( 1 + v ) 1 + v = ω 0 1 − v 1 + v (3.290) ω = ω 0 ( 1 − v ) ( 1 + v ) 1 + v = ω 0 1 − v 1 + v {:(3.290)omega=omega_(0)(sqrt((1-v)(1+v)))/(1+v)=omega_(0)sqrt((1-v)/(1+v)):}\begin{equation*} \omega=\omega_{0} \frac{\sqrt{(1-v)(1+v)}}{1+v}=\omega_{0} \sqrt{\frac{1-v}{1+v}} \tag{3.290} \end{equation*}(3.290)ω=ω0(1−v)(1+v)1+v=ω01−v1+v
b) Requiring that ω = ω 0 ω = ω 0 omega=omega_(0)\omega=\omega_{0}ω=ω0 results in
(3.291) 1 = 1 − v 2 1 + v cos ⁡ ( θ ) (3.291) 1 = 1 − v 2 1 + v cos ⁡ ( θ ) {:(3.291)1=(sqrt(1-v^(2)))/(1+v cos(theta)):}\begin{equation*} 1=\frac{\sqrt{1-v^{2}}}{1+v \cos (\theta)} \tag{3.291} \end{equation*}(3.291)1=1−v21+vcos⁡(θ)
Solving for cos ⁡ ( θ ) cos ⁡ ( θ ) cos(theta)\cos (\theta)cos⁡(θ) yields the result
(3.292) cos ⁡ ( θ ) = 1 − v 2 − 1 v (3.292) cos ⁡ ( θ ) = 1 − v 2 − 1 v {:(3.292)cos(theta)=(sqrt(1-v^(2))-1)/(v):}\begin{equation*} \cos (\theta)=\frac{\sqrt{1-v^{2}}-1}{v} \tag{3.292} \end{equation*}(3.292)cos⁡(θ)=1−v2−1v
Note that for v ≪ 1 v ≪ 1 v≪1v \ll 1v≪1, this may be approximated by
(3.293) cos ⁡ ( θ ) ≃ 1 − v 2 / 2 − 1 v = − v 2 (3.293) cos ⁡ ( θ ) ≃ 1 − v 2 / 2 − 1 v = − v 2 {:(3.293)cos(theta)≃(1-v^(2)//2-1)/(v)=-(v)/(2):}\begin{equation*} \cos (\theta) \simeq \frac{1-v^{2} / 2-1}{v}=-\frac{v}{2} \tag{3.293} \end{equation*}(3.293)cos⁡(θ)≃1−v2/2−1v=−v2
while for v = 1 − ε v = 1 − ε v=1-epsiv=1-\varepsilonv=1−ε, where ε ≪ 1 ε ≪ 1 epsi≪1\varepsilon \ll 1ε≪1, leads to
(3.294) cos ⁡ ( θ ) ≃ 1 − ( 1 − 2 ε ) − 1 1 − ε ≃ − 1 + 2 ε (3.294) cos ⁡ ( θ ) ≃ 1 − ( 1 − 2 ε ) − 1 1 − ε ≃ − 1 + 2 ε {:(3.294)cos(theta)≃(sqrt(1-(1-2epsi))-1)/(1-epsi)≃-1+sqrt(2epsi):}\begin{equation*} \cos (\theta) \simeq \frac{\sqrt{1-(1-2 \varepsilon)}-1}{1-\varepsilon} \simeq-1+\sqrt{2 \varepsilon} \tag{3.294} \end{equation*}(3.294)cos⁡(θ)≃1−(1−2ε)−11−ε≃−1+2ε
Thus, for small velocities, the angle θ θ theta\thetaθ is close to 90 ∘ 90 ∘ 90^(@)90^{\circ}90∘, but the source must move slightly toward the observer, whereas for velocities close to the speed of light, the source must move almost straight toward the observer.

1.66

We have a rotating disk with two observers O 1 O 1 O_(1)O_{1}O1 and O 2 O 2 O_(2)O_{2}O2 at different radii r 1 r 1 r_(1)r_{1}r1 and r 2 r 2 r_(2)r_{2}r2, respectively. Setting c = 1 c = 1 c=1c=1c=1 and omitting the trivial z z zzz-direction, introducing polar coordinates on the inertial frame leads to
(3.295) d s 2 = d t 2 − d r 2 − r 2 d ϕ 2 (3.295) d s 2 = d t 2 − d r 2 − r 2 d ϕ 2 {:(3.295)ds^(2)=dt^(2)-dr^(2)-r^(2)dphi^(2):}\begin{equation*} d s^{2}=d t^{2}-d r^{2}-r^{2} d \phi^{2} \tag{3.295} \end{equation*}(3.295)ds2=dt2−dr2−r2dϕ2
We now introduce a rotating frame with ϕ r = ϕ + Ω t ϕ r = ϕ + Ω t phi_(r)=phi+Omega t\phi_{r}=\phi+\Omega tϕr=ϕ+Ωt, which leads to ϕ = ϕ r − Ω t ϕ = ϕ r − Ω t phi=phi_(r)-Omega t\phi=\phi_{r}-\Omega tϕ=ϕr−Ωt and find that
d s 2 = d t 2 − d r 2 − r 2 d ϕ 2 = d t 2 − d r 2 − r 2 ( d ϕ r − Ω d t ) 2 (3.296) = ( 1 − r 2 Ω 2 ) d t 2 + r 2 Ω d t d ϕ r − d r 2 − r 2 d ϕ r 2 d s 2 = d t 2 − d r 2 − r 2 d ϕ 2 = d t 2 − d r 2 − r 2 d ϕ r − Ω d t 2 (3.296) = 1 − r 2 Ω 2 d t 2 + r 2 Ω d t d ϕ r − d r 2 − r 2 d ϕ r 2 {:[ds^(2)=dt^(2)-dr^(2)-r^(2)dphi^(2)=dt^(2)-dr^(2)-r^(2)(dphi_(r)-Omega dt)^(2)],[(3.296)=(1-r^(2)Omega^(2))dt^(2)+r^(2)Omega dtdphi_(r)-dr^(2)-r^(2)dphi_(r)^(2)]:}\begin{align*} d s^{2} & =d t^{2}-d r^{2}-r^{2} d \phi^{2}=d t^{2}-d r^{2}-r^{2}\left(d \phi_{r}-\Omega d t\right)^{2} \\ & =\left(1-r^{2} \Omega^{2}\right) d t^{2}+r^{2} \Omega d t d \phi_{r}-d r^{2}-r^{2} d \phi_{r}^{2} \tag{3.296} \end{align*}ds2=dt2−dr2−r2dϕ2=dt2−dr2−r2(dϕr−Ωdt)2(3.296)=(1−r2Ω2)dt2+r2Ωdtdϕr−dr2−r2dϕr2
Thus, we observe that we have a nontrivial time-time component of the metric, i.e.,
(3.297) g 00 = 1 − r 2 Ω 2 (3.297) g 00 = 1 − r 2 Ω 2 {:(3.297)g_(00)=1-r^(2)Omega^(2):}\begin{equation*} g_{00}=1-r^{2} \Omega^{2} \tag{3.297} \end{equation*}(3.297)g00=1−r2Ω2
For a light wave sent from O 2 O 2 O_(2)O_{2}O2 to O 1 O 1 O_(1)O_{1}O1, the observed proper times between consecutive wave fronts are therefore given by
(3.298) Δ τ 1 = g 00 ( r 1 ) Δ t , Δ τ 2 = g 00 ( r 2 ) Δ t (3.298) Δ τ 1 = g 00 r 1 Δ t , Δ τ 2 = g 00 r 2 Δ t {:(3.298)Deltatau_(1)=sqrt(g_(00)(r_(1)))Delta t","quad Deltatau_(2)=sqrt(g_(00)(r_(2)))Delta t:}\begin{equation*} \Delta \tau_{1}=\sqrt{g_{00}\left(r_{1}\right)} \Delta t, \quad \Delta \tau_{2}=\sqrt{g_{00}\left(r_{2}\right)} \Delta t \tag{3.298} \end{equation*}(3.298)Δτ1=g00(r1)Δt,Δτ2=g00(r2)Δt
where Δ t Δ t Delta t\Delta tΔt is the time difference between the emission of the wave fronts in the inertial frame. Therefore, using that the observed angular frequency ω ω omega\omegaω is inversely proportional to the time period, we obtain
(3.299) ω 2 ω 1 = Δ τ 1 Δ τ 2 = g 00 ( r 1 ) g 00 ( r 2 ) = 1 − r 1 2 Ω 2 1 − r 2 2 Ω 2 = γ 2 γ 1 (3.299) ω 2 ω 1 = Δ τ 1 Δ τ 2 = g 00 r 1 g 00 r 2 = 1 − r 1 2 Ω 2 1 − r 2 2 Ω 2 = γ 2 γ 1 {:(3.299)(omega_(2))/(omega_(1))=(Deltatau_(1))/(Deltatau_(2))=(sqrt(g_(00)(r_(1))))/(sqrt(g_(00)(r_(2))))=(sqrt(1-r_(1)^(2)Omega^(2)))/(sqrt(1-r_(2)^(2)Omega^(2)))=(gamma_(2))/(gamma_(1)):}\begin{equation*} \frac{\omega_{2}}{\omega_{1}}=\frac{\Delta \tau_{1}}{\Delta \tau_{2}}=\frac{\sqrt{g_{00}\left(r_{1}\right)}}{\sqrt{g_{00}\left(r_{2}\right)}}=\frac{\sqrt{1-r_{1}^{2} \Omega^{2}}}{\sqrt{1-r_{2}^{2} \Omega^{2}}}=\frac{\gamma_{2}}{\gamma_{1}} \tag{3.299} \end{equation*}(3.299)ω2ω1=Δτ1Δτ2=g00(r1)g00(r2)=1−r12Ω21−r22Ω2=γ2γ1

1.67

In the rest frame of the medium S S SSS, the waves in direction θ θ theta\thetaθ have a 4 -frequency given by
(3.300) ( N μ ) = v ( 1 , cos ⁡ ( θ ) / n , sin ⁡ ( θ ) / n ) (3.300) N μ = v ( 1 , cos ⁡ ( θ ) / n , sin ⁡ ( θ ) / n ) {:(3.300)(N^(mu))=v(1","cos(theta)//n","sin(theta)//n):}\begin{equation*} \left(N^{\mu}\right)=v(1, \cos (\theta) / n, \sin (\theta) / n) \tag{3.300} \end{equation*}(3.300)(Nμ)=v(1,cos⁡(θ)/n,sin⁡(θ)/n)
where the coordinate system has been arranged such that the third spatial component is zero (which is why it has been omitted). Here, v v vvv is the frequency of the wave in the rest frame of the medium. In the rest frame of the source S ′ S ′ S^(')S^{\prime}S′, the frequency is given by the zeroth component of the 4 -frequency in that frame
(3.301) v 0 = ( 1 , 0 , 0 ) μ N ′ μ = V μ ′ N ′ μ (3.301) v 0 = ( 1 , 0 , 0 ) μ N ′ μ = V μ ′ N ′ μ {:(3.301)v_(0)=(1","0","0)_(mu)N^('mu)=V_(mu)^(')N^('mu):}\begin{equation*} v_{0}=(1,0,0)_{\mu} N^{\prime \mu}=V_{\mu}^{\prime} N^{\prime \mu} \tag{3.301} \end{equation*}(3.301)v0=(1,0,0)μN′μ=Vμ′N′μ
where V V VVV is the 4 -velocity of the source itself. This expression is a Lorentz scalar and may be computed in any frame. In particular, in S S SSS, the 4 -velocity of the source is
(3.302) ( V μ ) = γ ( 1 , v , 0 ) , v 0 = V ⋅ N = γ v ( 1 − v cos ⁡ θ n ) (3.302) V μ = γ ( 1 , v , 0 ) , v 0 = V ⋅ N = γ v 1 − v cos ⁡ θ n {:(3.302)(V^(mu))=gamma(1","v","0)","quadv_(0)=V*N=gamma v(1-v(cos theta)/(n)):}\begin{equation*} \left(V^{\mu}\right)=\gamma(1, v, 0), \quad v_{0}=V \cdot N=\gamma v\left(1-v \frac{\cos \theta}{n}\right) \tag{3.302} \end{equation*}(3.302)(Vμ)=γ(1,v,0),v0=V⋅N=γv(1−vcos⁡θn)
It follows that
(3.303) v v 0 = 1 − v 2 n n − v cos ⁡ θ (3.303) v v 0 = 1 − v 2 n n − v cos ⁡ θ {:(3.303)(v)/(v_(0))=sqrt(1-v^(2))(n)/(n-v cos theta):}\begin{equation*} \frac{v}{v_{0}}=\sqrt{1-v^{2}} \frac{n}{n-v \cos \theta} \tag{3.303} \end{equation*}(3.303)vv0=1−v2nn−vcos⁡θ
In particular, when n → 1 n → 1 n rarr1n \rightarrow 1n→1, we recover
(3.304) v v 0 = 1 − v 2 1 − v cos ⁡ θ ⟶ θ → 0 1 + v 1 − v (3.304) v v 0 = 1 − v 2 1 − v cos ⁡ θ ⟶ θ → 0 1 + v 1 − v {:(3.304)(v)/(v_(0))=(sqrt(1-v^(2)))/(1-v cos theta)longrightarrow_(theta rarr0)sqrt((1+v)/(1-v)):}\begin{equation*} \frac{v}{v_{0}}=\frac{\sqrt{1-v^{2}}}{1-v \cos \theta} \underset{\theta \rightarrow 0}{\longrightarrow} \sqrt{\frac{1+v}{1-v}} \tag{3.304} \end{equation*}(3.304)vv0=1−v21−vcos⁡θ⟶θ→01+v1−v
which is the usual Doppler formula in vacuum.

1.68

In the frame S S SSS, where the mirror is moving, the 4 -frequency of the incoming light p p ppp and that of the outgoing light k k kkk are given by
(3.305) p = ω ( 1 , − c i , − s i ) , k = ω ′ ( 1 , c o , − s o ) , (3.305) p = ω 1 , − c i , − s i , k = ω ′ 1 , c o , − s o , {:(3.305)p=omega(1,-c_(i),-s_(i))","quad k=omega^(')(1,c_(o),-s_(o))",":}\begin{equation*} p=\omega\left(1,-c_{i},-s_{i}\right), \quad k=\omega^{\prime}\left(1, c_{o},-s_{o}\right), \tag{3.305} \end{equation*}(3.305)p=ω(1,−ci,−si),k=ω′(1,co,−so),
where we have introduced c i = cos ⁡ θ in , s i = sin ⁡ θ in , c o = cos ⁡ θ out , s o = sin ⁡ θ out c i = cos ⁡ θ in  , s i = sin ⁡ θ in  , c o = cos ⁡ θ out  , s o = sin ⁡ θ out  c_(i)=cos theta_("in "),s_(i)=sin theta_("in "),c_(o)=cos theta_("out "),s_(o)=sin theta_("out ")c_{i}=\cos \theta_{\text {in }}, s_{i}=\sin \theta_{\text {in }}, c_{o}=\cos \theta_{\text {out }}, s_{o}=\sin \theta_{\text {out }}ci=cos⁡θin ,si=sin⁡θin ,co=cos⁡θout ,so=sin⁡θout , and used a coordinate system such that the mirror is moving in the x x xxx-direction and the light is not propagating in the z z zzz-direction, which we therefore have omitted. In the rest frame of the mirror S ′ S ′ S^(')S^{\prime}S′, the incident angle is equal to the reflected angle. Furthermore, the frequencies of the incoming and reflected light are the same, which means that
(3.306) p ′ = ω ′ ′ ( 1 , − c ′ , − s ′ ) , k ′ = ω ′ ′ ( 1 , c ′ , − s ′ ) (3.306) p ′ = ω ′ ′ 1 , − c ′ , − s ′ , k ′ = ω ′ ′ 1 , c ′ , − s ′ {:(3.306)p^(')=omega^('')(1,-c^('),-s^('))","quadk^(')=omega^('')(1,c^('),-s^(')):}\begin{equation*} p^{\prime}=\omega^{\prime \prime}\left(1,-c^{\prime},-s^{\prime}\right), \quad k^{\prime}=\omega^{\prime \prime}\left(1, c^{\prime},-s^{\prime}\right) \tag{3.306} \end{equation*}(3.306)p′=ω′′(1,−c′,−s′),k′=ω′′(1,c′,−s′)
where c ′ = cos ⁡ θ ′ , s ′ = sin ⁡ θ ′ c ′ = cos ⁡ θ ′ , s ′ = sin ⁡ θ ′ c^(')=cos theta^('),s^(')=sin theta^(')c^{\prime}=\cos \theta^{\prime}, s^{\prime}=\sin \theta^{\prime}c′=cos⁡θ′,s′=sin⁡θ′. Since p p ppp and k k kkk are related to p ′ p ′ p^(')p^{\prime}p′ and k ′ k ′ k^(')k^{\prime}k′ by Lorentz transformation, we also have
(3.307) p ′ = ω γ ( 1 + v c i , − c i − v , s i / γ ) , k = ω ′ ′ γ ( 1 + v c ′ , c ′ + v , s ′ / γ ) (3.307) p ′ = ω γ 1 + v c i , − c i − v , s i / γ , k = ω ′ ′ γ 1 + v c ′ , c ′ + v , s ′ / γ {:(3.307)p^(')=omega gamma(1+vc_(i),-c_(i)-v,s_(i)//gamma)","quad k=omega^('')gamma(1+vc^('),c^(')+v,s^(')//gamma):}\begin{equation*} p^{\prime}=\omega \gamma\left(1+v c_{i},-c_{i}-v, s_{i} / \gamma\right), \quad k=\omega^{\prime \prime} \gamma\left(1+v c^{\prime}, c^{\prime}+v, s^{\prime} / \gamma\right) \tag{3.307} \end{equation*}(3.307)p′=ωγ(1+vci,−ci−v,si/γ),k=ω′′γ(1+vc′,c′+v,s′/γ)
where γ = 1 / 1 − v 2 γ = 1 / 1 − v 2 gamma=1//sqrt(1-v^(2))\gamma=1 / \sqrt{1-v^{2}}γ=1/1−v2. By identification, it follows (using p ′ p ′ p^(')p^{\prime}p′ ) that
(3.308) ω ′ ′ = ω γ ( 1 + v c i ) , c ′ = c i + v 1 + v c i (3.308) ω ′ ′ = ω γ 1 + v c i , c ′ = c i + v 1 + v c i {:(3.308)omega^('')=omega gamma(1+vc_(i))","quadc^(')=(c_(i)+v)/(1+vc_(i)):}\begin{equation*} \omega^{\prime \prime}=\omega \gamma\left(1+v c_{i}\right), \quad c^{\prime}=\frac{c_{i}+v}{1+v c_{i}} \tag{3.308} \end{equation*}(3.308)ω′′=ωγ(1+vci),c′=ci+v1+vci
and (using k k kkk ) that
(3.309) ω ′ = ω ′ ′ γ ( 1 + v c ′ ) = ω 1 + 2 v c i + v 2 1 − v 2 (3.310) c o = ω ′ ′ γ ω ′ ( c ′ + v ) = c ′ + v 1 + v c ′ = c i + 2 v + v 2 c i 1 + 2 v c i + v 2 (3.309) ω ′ = ω ′ ′ γ 1 + v c ′ = ω 1 + 2 v c i + v 2 1 − v 2 (3.310) c o = ω ′ ′ γ ω ′ c ′ + v = c ′ + v 1 + v c ′ = c i + 2 v + v 2 c i 1 + 2 v c i + v 2 {:[(3.309)omega^(')=omega^('')gamma(1+vc^('))=omega(1+2vc_(i)+v^(2))/(1-v^(2))],[(3.310)c_(o)=(omega^('')gamma)/(omega^('))(c^(')+v)=(c^(')+v)/(1+vc^('))=(c_(i)+2v+v^(2)c_(i))/(1+2vc_(i)+v^(2))]:}\begin{align*} & \omega^{\prime}=\omega^{\prime \prime} \gamma\left(1+v c^{\prime}\right)=\omega \frac{1+2 v c_{i}+v^{2}}{1-v^{2}} \tag{3.309}\\ & c_{o}=\frac{\omega^{\prime \prime} \gamma}{\omega^{\prime}}\left(c^{\prime}+v\right)=\frac{c^{\prime}+v}{1+v c^{\prime}}=\frac{c_{i}+2 v+v^{2} c_{i}}{1+2 v c_{i}+v^{2}} \tag{3.310} \end{align*}(3.309)ω′=ω′′γ(1+vc′)=ω1+2vci+v21−v2(3.310)co=ω′′γω′(c′+v)=c′+v1+vc′=ci+2v+v2ci1+2vci+v2
For v = − c i = − cos ⁡ θ in v = − c i = − cos ⁡ θ in v=-c_(i)=-cos theta_(in)v=-c_{i}=-\cos \theta_{\mathrm{in}}v=−ci=−cos⁡θin, the mirror is moving away from the light at the same speed that the light is approaching the mirror. As a result, the light never reaches the mirror. In S ′ S ′ S^(')S^{\prime}S′, the light is moving parallel to the mirror. The reflection angle in S S SSS approaches c o = − c i c o = − c i c_(o)=-c_(i)c_{o}=-c_{i}co=−ci, which simply means that the light continues in a straight line. For values v ≤ c i v ≤ c i v <= c_(i)v \leq c_{i}v≤ci, the mirror outruns the light and there is no reflection.

1.69

In the rest frame S ′ S ′ S^(')S^{\prime}S′ of the medium, the 4 -frequencies of the light waves are given by
(3.311) N μ ′ = f 0 ( 1 , − c ′ , − s ′ ) μ ′ , N μ ′ = f 0 ( 1 , − c ′ ′ n , − s ′ ′ n ) μ ′ (3.311) N μ ′ = f 0 1 , − c ′ , − s ′ μ ′ , N μ ′ = f 0 1 , − c ′ ′ n , − s ′ ′ n μ ′ {:(3.311)N^(mu^('))=f_(0)(1,-c^('),-s^('))^(mu^('))","quadN^(mu^('))=f_(0)(1,-(c^(''))/(n),-(s^(''))/(n))^(mu^(')):}\begin{equation*} N^{\mu^{\prime}}=f_{0}\left(1,-c^{\prime},-s^{\prime}\right)^{\mu^{\prime}}, \quad \mathcal{N}^{\mu^{\prime}}=f_{0}\left(1,-\frac{c^{\prime \prime}}{n},-\frac{s^{\prime \prime}}{n}\right)^{\mu^{\prime}} \tag{3.311} \end{equation*}(3.311)Nμ′=f0(1,−c′,−s′)μ′,Nμ′=f0(1,−c′′n,−s′′n)μ′
where c ′ = cos ⁡ θ ′ , s ′ = sin ⁡ θ ′ , s ′ ′ = sin ⁡ θ ′ ′ , c ′ ′ = cos ⁡ θ ′ ′ c ′ = cos ⁡ θ ′ , s ′ = sin ⁡ θ ′ , s ′ ′ = sin ⁡ θ ′ ′ , c ′ ′ = cos ⁡ θ ′ ′ c^(')=cos theta^('),s^(')=sin theta^('),s^('')=sin theta^(''),c^('')=cos theta^('')c^{\prime}=\cos \theta^{\prime}, s^{\prime}=\sin \theta^{\prime}, s^{\prime \prime}=\sin \theta^{\prime \prime}, c^{\prime \prime}=\cos \theta^{\prime \prime}c′=cos⁡θ′,s′=sin⁡θ′,s′′=sin⁡θ′′,c′′=cos⁡θ′′ and the relation between θ ′ θ ′ theta^(')\theta^{\prime}θ′ and θ ′ ′ θ ′ ′ theta^('')\theta^{\prime \prime}θ′′ is given by Snell's law sin ⁡ θ ′ = n sin ⁡ θ ′ ′ sin ⁡ θ ′ = n sin ⁡ θ ′ ′ sin theta^(')=n sin theta^('')\sin \theta^{\prime}=n \sin \theta^{\prime \prime}sin⁡θ′=nsin⁡θ′′. Here N N NNN is the 4 -frequency of the wave before entering the medium and N N N\mathcal{N}N is the 4 -frequency after entering the medium. Expressing θ ′ ′ θ ′ ′ theta^('')\theta^{\prime \prime}θ′′ in terms of θ ′ θ ′ theta^(')\theta^{\prime}θ′, we obtain
(3.312) N μ ′ = f 0 ( 1 , − 1 n 1 − s ′ 2 n 2 , − s ′ n 2 ) μ ′ (3.312) N μ ′ = f 0 1 , − 1 n 1 − s ′ 2 n 2 , − s ′ n 2 μ ′ {:(3.312)N^(mu^('))=f_(0)(1,-(1)/(n)sqrt(1-(s^('2))/(n^(2))),-(s^('))/(n^(2)))^(mu^(')):}\begin{equation*} N^{\mu^{\prime}}=f_{0}\left(1,-\frac{1}{n} \sqrt{1-\frac{s^{\prime 2}}{n^{2}}},-\frac{s^{\prime}}{n^{2}}\right)^{\mu^{\prime}} \tag{3.312} \end{equation*}(3.312)Nμ′=f0(1,−1n1−s′2n2,−s′n2)μ′
Lorentz transforming this to the frame S S SSS, we obtain
(3.313) N μ = f 0 γ ( 1 − v c ′ , − c ′ + v , − s ′ γ ) μ (3.314) N μ = f 0 γ ( 1 − v n 1 − s ′ 2 n 2 , + v − 1 n 1 − s ′ 2 n 2 , − s ′ n 2 γ ) μ (3.313) N μ = f 0 γ 1 − v c ′ , − c ′ + v , − s ′ γ μ (3.314) N μ = f 0 γ 1 − v n 1 − s ′ 2 n 2 , + v − 1 n 1 − s ′ 2 n 2 , − s ′ n 2 γ μ {:[(3.313)N^(mu)=f_(0)gamma(1-vc^('),-c^(')+v,-(s^('))/(gamma))^(mu)],[(3.314)N^(mu)=f_(0)gamma(1-(v)/(n)sqrt(1-(s^('2))/(n^(2))),+v-(1)/(n)sqrt(1-(s^('2))/(n^(2))),-(s^('))/(n^(2)gamma))^(mu)]:}\begin{align*} & N^{\mu}=f_{0} \gamma\left(1-v c^{\prime},-c^{\prime}+v,-\frac{s^{\prime}}{\gamma}\right)^{\mu} \tag{3.313}\\ & \mathcal{N}^{\mu}=f_{0} \gamma\left(1-\frac{v}{n} \sqrt{1-\frac{s^{\prime 2}}{n^{2}}},+v-\frac{1}{n} \sqrt{1-\frac{s^{\prime 2}}{n^{2}}},-\frac{s^{\prime}}{n^{2} \gamma}\right)^{\mu} \tag{3.314} \end{align*}(3.313)Nμ=f0γ(1−vc′,−c′+v,−s′γ)μ(3.314)Nμ=f0γ(1−vn1−s′2n2,+v−1n1−s′2n2,−s′n2γ)μ
In order for the observer in S S SSS to observe the light traveling in the same direction after entering the medium, the relation
(3.315) N 1 N 1 = N 2 N 2 (3.315) N 1 N 1 = N 2 N 2 {:(3.315)(N^(1))/(N^(1))=(N^(2))/(N^(2)):}\begin{equation*} \frac{\mathcal{N}^{1}}{N^{1}}=\frac{\mathcal{N}^{2}}{N^{2}} \tag{3.315} \end{equation*}(3.315)N1N1=N2N2
must be fulfilled (the spatial part of the 4 -frequencies must be in the same direction). We obtain
(3.316) N 1 N 1 = 1 n 1 − s ′ 2 n 2 − v c ′ − v (3.317) N 2 N 2 = s ′ n 2 γ s ′ γ = 1 n 2 (3.316) N 1 N 1 = 1 n 1 − s ′ 2 n 2 − v c ′ − v (3.317) N 2 N 2 = s ′ n 2 γ s ′ γ = 1 n 2 {:[(3.316)(N^(1))/(N^(1))=((1)/(n)sqrt(1-(s^('2))/(n^(2)))-v)/(c^(')-v)],[(3.317)(N^(2))/(N^(2))=((s^('))/(n^(2)gamma))/((s^('))/(gamma))=(1)/(n^(2))]:}\begin{align*} & \frac{\mathcal{N}^{1}}{N^{1}}=\frac{\frac{1}{n} \sqrt{1-\frac{s^{\prime 2}}{n^{2}}}-v}{c^{\prime}-v} \tag{3.316}\\ & \frac{\mathcal{N}^{2}}{N^{2}}=\frac{\frac{s^{\prime}}{n^{2} \gamma}}{\frac{s^{\prime}}{\gamma}}=\frac{1}{n^{2}} \tag{3.317} \end{align*}(3.316)N1N1=1n1−s′2n2−vc′−v(3.317)N2N2=s′n2γs′γ=1n2
This leads to the relation
(3.318) c ′ − v = n 2 − s ′ 2 − n 2 v ⇒ c ′ + v ( n 2 − 1 ) = n 2 − 1 + c ′ 2 (3.318) c ′ − v = n 2 − s ′ 2 − n 2 v ⇒ c ′ + v n 2 − 1 = n 2 − 1 + c ′ 2 {:(3.318)c^(')-v=sqrt(n^(2)-s^('2))-n^(2)v quad=>quadc^(')+v(n^(2)-1)=sqrt(n^(2)-1+c^('2)):}\begin{equation*} c^{\prime}-v=\sqrt{n^{2}-s^{\prime 2}}-n^{2} v \quad \Rightarrow \quad c^{\prime}+v\left(n^{2}-1\right)=\sqrt{n^{2}-1+c^{\prime 2}} \tag{3.318} \end{equation*}(3.318)c′−v=n2−s′2−n2v⇒c′+v(n2−1)=n2−1+c′2
Squaring this, we obtain
(3.319) ( n 2 − 1 ) [ 2 c ′ v − ( 1 − v 2 ) ( n 2 − 1 ) ] = 0 (3.319) n 2 − 1 2 c ′ v − 1 − v 2 n 2 − 1 = 0 {:(3.319)(n^(2)-1)[2c^(')v-(1-v^(2))(n^(2)-1)]=0:}\begin{equation*} \left(n^{2}-1\right)\left[2 c^{\prime} v-\left(1-v^{2}\right)\left(n^{2}-1\right)\right]=0 \tag{3.319} \end{equation*}(3.319)(n2−1)[2c′v−(1−v2)(n2−1)]=0
which has the solutions n 2 = 1 n 2 = 1 n^(2)=1n^{2}=1n2=1 and n 2 = 1 + 2 c ′ v γ 2 n 2 = 1 + 2 c ′ v γ 2 n^(2)=1+2c^(')vgamma^(2)n^{2}=1+2 c^{\prime} v \gamma^{2}n2=1+2c′vγ2. Since n > 1 n > 1 n > 1n>1n>1 was given, the latter of these is the sought solution and we obtain
(3.320) n = 1 + 2 c ′ v γ 2 (3.320) n = 1 + 2 c ′ v γ 2 {:(3.320)n=sqrt(1+2c^(')vgamma^(2)):}\begin{equation*} n=\sqrt{1+2 c^{\prime} v \gamma^{2}} \tag{3.320} \end{equation*}(3.320)n=1+2c′vγ2
Note that the solution n = 1 n = 1 n=1n=1n=1 corresponds to the scenario where the medium has the same refractive index as vacuum, i.e., is vacuum. In this case, it does not matter how fast the medium moves and the light wave will continue undisturbed.

1.70

We can find the angular frequency ω 0 ω 0 omega_(0)\omega_{0}ω0 of the wave in the source rest frame by multiplying the 4 -frequency N N NNN by the 4 -velocity of the source. This results in
(3.321) ω 0 = N ⋅ V = γ ( ω , k ) ⋅ ( 1 , v ) = γ ω ( 1 − v u ) (3.321) ω 0 = N ⋅ V = γ ( ω , k ) ⋅ ( 1 , v ) = γ ω 1 − v u {:(3.321)omega_(0)=N*V=gamma(omega","k)*(1","v)=gamma omega(1-(v)/(u)):}\begin{equation*} \omega_{0}=N \cdot V=\gamma(\omega, k) \cdot(1, v)=\gamma \omega\left(1-\frac{v}{u}\right) \tag{3.321} \end{equation*}(3.321)ω0=N⋅V=γ(ω,k)⋅(1,v)=γω(1−vu)
Solving for the Doppler shifted frequency ω ω omega\omegaω results in
(3.322) ω = ω 0 u ( u − v ) γ (3.322) ω = ω 0 u ( u − v ) γ {:(3.322)omega=(omega_(0)u)/((u-v)gamma):}\begin{equation*} \omega=\frac{\omega_{0} u}{(u-v) \gamma} \tag{3.322} \end{equation*}(3.322)ω=ω0u(u−v)γ
When v → u v → u v rarr uv \rightarrow uv→u, the source is moving essentially at the same velocity as the wave speed in the medium and in the same direction as the wave. As a result, the wave train is compressed and the frequency approaches infinity. This is the same behavior as obtained for the classical Doppler shift. When v → − u v → − u v rarr-uv \rightarrow-uv→−u, the source moves in the opposite direction to the wave, resulting in a frequency ω → ω 0 / ( 2 γ ) ω → ω 0 / ( 2 γ ) omega rarromega_(0)//(2gamma)\omega \rightarrow \omega_{0} /(2 \gamma)ω→ω0/(2γ). The γ γ gamma\gammaγ factor appears due to the time dilation of the source, while the factor of two results from the wavelength being doubled due to the motion of the source. The only difference here to the classical Doppler shift is the appearance of the factor γ γ gamma\gammaγ, describing the time dilation due to the motion of the source.

1.71

From the form of the worldline for observer B B BBB, we find that
(3.323) x x ˙ − t t ˙ = 0 (3.323) x x ˙ − t t ˙ = 0 {:(3.323)xx^(˙)-tt^(˙)=0:}\begin{equation*} x \dot{x}-t \dot{t}=0 \tag{3.323} \end{equation*}(3.323)xx˙−tt˙=0
by differentiating with respect to the proper time. It directly follows that v = v = v=v=v= x ˙ / t ˙ = t / x x ˙ / t ˙ = t / x x^(˙)//t^(˙)=t//x\dot{x} / \dot{t}=t / xx˙/t˙=t/x. Solving for x x xxx in terms of t t ttt and inserting it into this expression now gives
(3.324) v 2 = α 2 t 2 1 + α 2 t 2 (3.324) v 2 = α 2 t 2 1 + α 2 t 2 {:(3.324)v^(2)=(alpha^(2)t^(2))/(1+alpha^(2)t^(2)):}\begin{equation*} v^{2}=\frac{\alpha^{2} t^{2}}{1+\alpha^{2} t^{2}} \tag{3.324} \end{equation*}(3.324)v2=α2t21+α2t2
The 4 -frequency of the signal is given by Ω = ω ( 1 , 1 ) Ω = ω ( 1 , 1 ) Omega=omega(1,1)\Omega=\omega(1,1)Ω=ω(1,1) in S S SSS and the frequency observed by B B BBB can be found by taking the inner product of this 4 -frequency with the 4 -velocity of B B BBB. We find that
(3.325) ω B = Ω ⋅ V B = ω γ ( 1 − v ) (3.325) ω B = Ω ⋅ V B = ω γ ( 1 − v ) {:(3.325)omega_(B)=Omega*V_(B)=omega gamma(1-v):}\begin{equation*} \omega_{B}=\Omega \cdot V_{B}=\omega \gamma(1-v) \tag{3.325} \end{equation*}(3.325)ωB=Ω⋅VB=ωγ(1−v)
Inserting the expression we have found for v v vvv into this results in
(3.326) ω B = ω ( 1 + α 2 t 2 − α t ) (3.326) ω B = ω 1 + α 2 t 2 − α t {:(3.326)omega_(B)=omega(sqrt(1+alpha^(2)t^(2))-alpha t):}\begin{equation*} \omega_{B}=\omega\left(\sqrt{1+\alpha^{2} t^{2}}-\alpha t\right) \tag{3.326} \end{equation*}(3.326)ωB=ω(1+α2t2−αt)
In general, this expression can also be found as
(3.327) ω B = ω 1 − v 1 + v (3.327) ω B = ω 1 − v 1 + v {:(3.327)omega_(B)=omegasqrt((1-v)/(1+v)):}\begin{equation*} \omega_{B}=\omega \sqrt{\frac{1-v}{1+v}} \tag{3.327} \end{equation*}(3.327)ωB=ω1−v1+v
while the frequency ω A ω A omega_(A)\omega_{A}ωA of the reflected signal is given by
(3.328) ω A = ω 1 − v 1 + v (3.328) ω A = ω 1 − v 1 + v {:(3.328)omega_(A)=omega(1-v)/(1+v):}\begin{equation*} \omega_{A}=\omega \frac{1-v}{1+v} \tag{3.328} \end{equation*}(3.328)ωA=ω1−v1+v
Direct comparison therefore results in
(3.329) ω A = ω ( 1 + α 2 t 2 − α t ) 2 (3.329) ω A = ω 1 + α 2 t 2 − α t 2 {:(3.329)omega_(A)=omega(sqrt(1+alpha^(2)t^(2))-alpha t)^(2):}\begin{equation*} \omega_{A}=\omega\left(\sqrt{1+\alpha^{2} t^{2}}-\alpha t\right)^{2} \tag{3.329} \end{equation*}(3.329)ωA=ω(1+α2t2−αt)2

1.72

The rest energy of an electron is E 0 = m 0 c 2 ≃ 0.51 MeV E 0 = m 0 c 2 ≃ 0.51 MeV E_(0)=m_(0)c^(2)≃0.51MeVE_{0}=m_{0} c^{2} \simeq 0.51 \mathrm{MeV}E0=m0c2≃0.51MeV, where m 0 m 0 m_(0)m_{0}m0 is the rest mass of the electron. Thus, the total energy after acceleration is therefore given by
(3.330) E = m 0 c 2 1 − v 2 c 2 = E 0 1 − v 2 c 2 = E 0 + 1 MeV ≃ 1.51 MeV (3.330) E = m 0 c 2 1 − v 2 c 2 = E 0 1 − v 2 c 2 = E 0 + 1 MeV ≃ 1.51 MeV {:(3.330)E=(m_(0)c^(2))/(sqrt(1-(v^(2))/(c^(2))))=(E_(0))/(sqrt(1-(v^(2))/(c^(2))))=E_(0)+1MeV≃1.51MeV:}\begin{equation*} E=\frac{m_{0} c^{2}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\frac{E_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=E_{0}+1 \mathrm{MeV} \simeq 1.51 \mathrm{MeV} \tag{3.330} \end{equation*}(3.330)E=m0c21−v2c2=E01−v2c2=E0+1MeV≃1.51MeV
Solving for v v vvv, we obtain
(3.331) v = c 1 − ( E 0 E ) 2 = c 1 − ( 0.51 1.51 ) 2 ≃ 0.94 c (3.331) v = c 1 − E 0 E 2 = c 1 − 0.51 1.51 2 ≃ 0.94 c {:(3.331)v=csqrt(1-((E_(0))/(E))^(2))=csqrt(1-((0.51)/(1.51))^(2))≃0.94 c:}\begin{equation*} v=c \sqrt{1-\left(\frac{E_{0}}{E}\right)^{2}}=c \sqrt{1-\left(\frac{0.51}{1.51}\right)^{2}} \simeq 0.94 c \tag{3.331} \end{equation*}(3.331)v=c1−(E0E)2=c1−(0.511.51)2≃0.94c
i.e., the final velocity of the electron is v ≃ 0.94 c v ≃ 0.94 c v≃0.94 cv \simeq 0.94 cv≃0.94c.

1.73

a) If the equation p e − + p e + = k γ p e − + p e + = k γ p_(e^(-))+p_(e^(+))=k_(gamma)p_{e^{-}}+p_{e^{+}}=k_{\gamma}pe−+pe+=kγ i.e., conservation of 4-momentum (where p e − p e − p_(e^(-))p_{e^{-}}pe−is the 4 -momentum of the electron, p e + p e + p_(e^(+))p_{e^{+}}pe+is the 4 -momentum of the positron, and k γ k γ k_(gamma)k_{\gamma}kγ is the 4 -momentum of the photon), is squared and the left-hand side is calculated in the rest frame of the electron, then the relation
(3.332) 2 m e 2 + 2 m e E e + = 0 (3.332) 2 m e 2 + 2 m e E e + = 0 {:(3.332)2m_(e)^(2)+2m_(e)E_(e^(+))=0:}\begin{equation*} 2 m_{e}^{2}+2 m_{e} E_{e^{+}}=0 \tag{3.332} \end{equation*}(3.332)2me2+2meEe+=0
is obtained, where m e m e m_(e)m_{e}me is the rest mass of the electron (or positron) and E e + E e + E_(e^(+))E_{e^{+}}Ee+is the total energy of the positron relative to the rest frame of the electron. This proves that this process is incompatible with conservation of 4 -momentum (i.e., conservation of energy and momentum) as all of the quantities on the left-hand side are strictly positive.
Alternatively, this can also be seen in an inertial frame where the spatial parts of the total momenta (i.e., the 3-momenta) of the electron and the positron are zero before the collision. In this frame, the 4 -momenta of the electron and the positron before the collision are
(3.333) p e − = ( m e 2 + p 2 , p ) and p e + = ( m e 2 + p 2 , − p ) (3.333) p e − = m e 2 + p 2 , p  and  p e + = m e 2 + p 2 , − p {:(3.333)p_(e^(-))=(sqrt(m_(e)^(2)+p^(2)),p)quad" and "quadp_(e^(+))=(sqrt(m_(e)^(2)+p^(2)),-p):}\begin{equation*} p_{e^{-}}=\left(\sqrt{m_{e}^{2}+\boldsymbol{p}^{2}}, \boldsymbol{p}\right) \quad \text { and } \quad p_{e^{+}}=\left(\sqrt{m_{e}^{2}+p^{2}},-\boldsymbol{p}\right) \tag{3.333} \end{equation*}(3.333)pe−=(me2+p2,p) and pe+=(me2+p2,−p)
where p = ( p 1 , p 2 , p 3 ) p = p 1 , p 2 , p 3 p=(p_(1),p_(2),p_(3))\boldsymbol{p}=\left(p_{1}, p_{2}, p_{3}\right)p=(p1,p2,p3). Let k γ = ( | k | , k ) k γ = ( | k | , k ) k_(gamma)=(|k|,k)k_{\gamma}=(|\boldsymbol{k}|, \boldsymbol{k})kγ=(|k|,k) be the lightlike 4-momentum of the photon. Then, conservation of 4-momentum p e − + p e + = k γ p e − + p e + = k γ p_(e^(-))+p_(e^(+))=k_(gamma)p_{e^{-}}+p_{e^{+}}=k_{\gamma}pe−+pe+=kγ implies that
(3.334) 2 m e 2 + p 2 = | k | and 0 = k (3.334) 2 m e 2 + p 2 = | k |  and  0 = k {:(3.334)2sqrt(m_(e)^(2)+p^(2))=|k|quad" and "quad0=k:}\begin{equation*} 2 \sqrt{m_{e}^{2}+p^{2}}=|k| \quad \text { and } \quad 0=k \tag{3.334} \end{equation*}(3.334)2me2+p2=|k| and 0=k
These two conditions clearly contradict each other.
b) Now, let p in p in  p_("in ")p_{\text {in }}pin  and p out p out  p_("out ")p_{\text {out }}pout  be the 4 -momentum of an electron before and after emitting a photon, respectively. In addition, let the photon have 4-momentum k k kkk. For this process to conserve the total 4-momentum, the relation
(3.335) p in = p out + k , (3.335) p in  = p out  + k , {:(3.335)p_("in ")=p_("out ")+k",":}\begin{equation*} p_{\text {in }}=p_{\text {out }}+k, \tag{3.335} \end{equation*}(3.335)pin =pout +k,
must hold. However, squaring this expression gives
(3.336) m e 2 = m e 2 + 2 p out ⋅ k = m e 2 + 2 m e ω , (3.336) m e 2 = m e 2 + 2 p out ⋅ k = m e 2 + 2 m e ω , {:(3.336)m_(e)^(2)=m_(e)^(2)+2p_(out)*k=m_(e)^(2)+2m_(e)omega",":}\begin{equation*} m_{e}^{2}=m_{e}^{2}+2 p_{\mathrm{out}} \cdot k=m_{e}^{2}+2 m_{e} \omega, \tag{3.336} \end{equation*}(3.336)me2=me2+2pout⋅k=me2+2meω,
where ω ω omega\omegaω is the photon energy in the rest frame of the electron after emitting the photon. This cannot hold for any nonzero photon energy ω ω omega\omegaω.
c) In this case, we have p μ + p μ ′ = k μ + k μ ′ p μ + p μ ′ = k μ + k μ ′ p_(mu)+p_(mu)^(')=k_(mu)+k_(mu)^(')p_{\mu}+p_{\mu}^{\prime}=k_{\mu}+k_{\mu}^{\prime}pμ+pμ′=kμ+kμ′, where ( k μ ) = ( | k | , k ) k μ = ( | k | , k ) (k_(mu))=(|k|,k)\left(k_{\mu}\right)=(|\boldsymbol{k}|, \boldsymbol{k})(kμ)=(|k|,k) and ( k μ ′ ) = k μ ′ = (k_(mu)^('))=\left(k_{\mu}^{\prime}\right)=(kμ′)= ( | k ′ | , k ′ ) k ′ , k ′ (|k^(')|,k^('))\left(\left|\boldsymbol{k}^{\prime}\right|, \boldsymbol{k}^{\prime}\right)(|k′|,k′) are the 4-momenta of the two photons, respectively. This implies that
(3.337) 2 m e 2 + p 2 = | k | + | k ′ | and 0 = k + k ′ (3.337) 2 m e 2 + p 2 = | k | + k ′  and  0 = k + k ′ {:(3.337)2sqrt(m_(e)^(2)+p^(2))=|k|+|k^(')|quad" and "quad0=k+k^('):}\begin{equation*} 2 \sqrt{m_{e}^{2}+p^{2}}=|\boldsymbol{k}|+\left|\boldsymbol{k}^{\prime}\right| \quad \text { and } \quad 0=\boldsymbol{k}+\boldsymbol{k}^{\prime} \tag{3.337} \end{equation*}(3.337)2me2+p2=|k|+|k′| and 0=k+k′
which clearly has a nontrivial solution k ′ = − k k ′ = − k k^(')=-kk^{\prime}=-\boldsymbol{k}k′=−k and | k | = m e 2 + p 2 | k | = m e 2 + p 2 |k|=sqrt(m_(e)^(2)+p^(2))|\boldsymbol{k}|=\sqrt{m_{e}^{2}+p^{2}}|k|=me2+p2. Thus, the answer to the question is "yes."

1.74

It holds that M 2 = P before 2 = P after 2 = ( p a + p b ) 2 M 2 = P before  2 = P after  2 = p a + p b 2 M^(2)=P_("before ")^(2)=P_("after ")^(2)=(p_(a)+p_(b))^(2)M^{2}=P_{\text {before }}^{2}=P_{\text {after }}^{2}=\left(p_{a}+p_{b}\right)^{2}M2=Pbefore 2=Pafter 2=(pa+pb)2, where p a = ( E a , p a ) p a = E a , p a p_(a)=(E_(a),p_(a))p_{a}=\left(E_{a}, \boldsymbol{p}_{a}\right)pa=(Ea,pa) and p b = p b = p_(b)=p_{b}=pb= ( m b , 0 ) m b , 0 (m_(b),0)\left(m_{b}, 0\right)(mb,0) in the rest frame of b b bbb. Solving this equation for | p a | p a |p_(a)|\left|p_{a}\right||pa|, using E a = m a 2 + p a 2 E a = m a 2 + p a 2 E_(a)=sqrt(m_(a)^(2)+p_(a)^(2))E_{a}=\sqrt{m_{a}^{2}+p_{a}^{2}}Ea=ma2+pa2, gives
(3.338) | p a | = 1 2 m b ⋅ [ M 2 − ( m a − m b ) 2 ] [ M 2 − ( m a + m b ) 2 ] (3.338) p a = 1 2 m b ⋅ M 2 − m a − m b 2 M 2 − m a + m b 2 {:(3.338)|p_(a)|=(1)/(2m_(b))*sqrt([M^(2)-(m_(a)-m_(b))^(2)][M^(2)-(m_(a)+m_(b))^(2)]):}\begin{equation*} \left|\boldsymbol{p}_{a}\right|=\frac{1}{2 m_{b}} \cdot \sqrt{\left[M^{2}-\left(m_{a}-m_{b}\right)^{2}\right]\left[M^{2}-\left(m_{a}+m_{b}\right)^{2}\right]} \tag{3.338} \end{equation*}(3.338)|pa|=12mb⋅[M2−(ma−mb)2][M2−(ma+mb)2]
1.75
Consider the reaction A ⟶ B + C A ⟶ B + C A longrightarrow B+CA \longrightarrow B+CA⟶B+C. Conservation of 4-momentum yields
(3.339) P A = P B + P C (3.339) P A = P B + P C {:(3.339)P_(A)=P_(B)+P_(C):}\begin{equation*} P_{A}=P_{B}+P_{C} \tag{3.339} \end{equation*}(3.339)PA=PB+PC
The redundant information about particle C C CCC can be removed by rewriting this as
(3.340) P C = P A − P B (3.340) P C = P A − P B {:(3.340)P_(C)=P_(A)-P_(B):}\begin{equation*} P_{C}=P_{A}-P_{B} \tag{3.340} \end{equation*}(3.340)PC=PA−PB
and squaring both sides, leading to
(3.341) P C 2 = ( P A − P B ) 2 = P A 2 + P B 2 − 2 P A ⋅ P B (3.341) P C 2 = P A − P B 2 = P A 2 + P B 2 − 2 P A ⋅ P B {:(3.341)P_(C)^(2)=(P_(A)-P_(B))^(2)=P_(A)^(2)+P_(B)^(2)-2P_(A)*P_(B):}\begin{equation*} P_{C}^{2}=\left(P_{A}-P_{B}\right)^{2}=P_{A}^{2}+P_{B}^{2}-2 P_{A} \cdot P_{B} \tag{3.341} \end{equation*}(3.341)PC2=(PA−PB)2=PA2+PB2−2PA⋅PB
Using the fact that P 2 = m 2 P 2 = m 2 P^(2)=m^(2)P^{2}=m^{2}P2=m2 we find
(3.342) m C 2 = m A 2 + m B 2 − 2 P A ⋅ P B (3.342) m C 2 = m A 2 + m B 2 − 2 P A ⋅ P B {:(3.342)m_(C)^(2)=m_(A)^(2)+m_(B)^(2)-2P_(A)*P_(B):}\begin{equation*} m_{C}^{2}=m_{A}^{2}+m_{B}^{2}-2 P_{A} \cdot P_{B} \tag{3.342} \end{equation*}(3.342)mC2=mA2+mB2−2PA⋅PB
In the rest frame of particle B B BBB (i.e., p B = 0 p B = 0 p_(B)=0\boldsymbol{p}_{B}=0pB=0 ), it holds that
(3.343) P A = ( E A , p A ) and P B = ( m B , 0 ) (3.343) P A = E A , p A  and  P B = m B , 0 {:(3.343)P_(A)=(E_(A),p_(A))quad" and "quadP_(B)=(m_(B),0):}\begin{equation*} P_{A}=\left(E_{A}, \boldsymbol{p}_{A}\right) \quad \text { and } \quad P_{B}=\left(m_{B}, \mathbf{0}\right) \tag{3.343} \end{equation*}(3.343)PA=(EA,pA) and PB=(mB,0)
which gives
(3.344) m C 2 = m A 2 + m B 2 − 2 E A m B (3.344) m C 2 = m A 2 + m B 2 − 2 E A m B {:(3.344)m_(C)^(2)=m_(A)^(2)+m_(B)^(2)-2E_(A)m_(B):}\begin{equation*} m_{C}^{2}=m_{A}^{2}+m_{B}^{2}-2 E_{A} m_{B} \tag{3.344} \end{equation*}(3.344)mC2=mA2+mB2−2EAmB
Using E A = m A 2 + p A 2 E A = m A 2 + p A 2 E_(A)=sqrt(m_(A)^(2)+p_(A)^(2))E_{A}=\sqrt{m_{A}^{2}+p_{A}^{2}}EA=mA2+pA2 and rearranging the above equation, we find that
(3.345) p A 2 = ( m A 2 + m B 2 − m C 2 2 m B ) 2 − m A 2 (3.345) p A 2 = m A 2 + m B 2 − m C 2 2 m B 2 − m A 2 {:(3.345)p_(A)^(2)=((m_(A)^(2)+m_(B)^(2)-m_(C)^(2))/(2m_(B)))^(2)-m_(A)^(2):}\begin{equation*} \boldsymbol{p}_{A}^{2}=\left(\frac{m_{A}^{2}+m_{B}^{2}-m_{C}^{2}}{2 m_{B}}\right)^{2}-m_{A}^{2} \tag{3.345} \end{equation*}(3.345)pA2=(mA2+mB2−mC22mB)2−mA2
Particle A A AAA has speed v A v A v_(A)v_{A}vA before the decay (relative to the rest frame of particle B B BBB after the decay), which means that
(3.346) p A = m v A = m A 1 − v A 2 v A (3.346) p A = m v A = m A 1 − v A 2 v A {:(3.346)p_(A)=mv_(A)=(m_(A))/(sqrt(1-v_(A)^(2)))v_(A):}\begin{equation*} \boldsymbol{p}_{A}=m \boldsymbol{v}_{A}=\frac{m_{A}}{\sqrt{1-v_{A}^{2}}} \boldsymbol{v}_{A} \tag{3.346} \end{equation*}(3.346)pA=mvA=mA1−vA2vA
i.e.,
(3.347) p A 2 = m A 2 v A 2 1 − v A 2 (3.347) p A 2 = m A 2 v A 2 1 − v A 2 {:(3.347)p_(A)^(2)=(m_(A)^(2)v_(A)^(2))/(1-v_(A)^(2)):}\begin{equation*} p_{A}^{2}=\frac{m_{A}^{2} v_{A}^{2}}{1-v_{A}^{2}} \tag{3.347} \end{equation*}(3.347)pA2=mA2vA21−vA2
Combining the two expressions for p A 2 p A 2 p_(A)^(2)\boldsymbol{p}_{A}^{2}pA2 yields
(3.348) m A 2 v A 2 1 − v A 2 = ( m A 2 + m B 2 − m C 2 2 m B ) 2 − m A 2 (3.348) m A 2 v A 2 1 − v A 2 = m A 2 + m B 2 − m C 2 2 m B 2 − m A 2 {:(3.348)(m_(A)^(2)v_(A)^(2))/(1-v_(A)^(2))=((m_(A)^(2)+m_(B)^(2)-m_(C)^(2))/(2m_(B)))^(2)-m_(A)^(2):}\begin{equation*} \frac{m_{A}^{2} v_{A}^{2}}{1-v_{A}^{2}}=\left(\frac{m_{A}^{2}+m_{B}^{2}-m_{C}^{2}}{2 m_{B}}\right)^{2}-m_{A}^{2} \tag{3.348} \end{equation*}(3.348)mA2vA21−vA2=(mA2+mB2−mC22mB)2−mA2
Solving for v A 2 v A 2 v_(A)^(2)v_{A}^{2}vA2, we obtain
(3.349) v A 2 = m A 4 + m B 4 + m C 4 − 2 m A 2 m B 2 − 2 m A 2 m C 2 − 2 m B 2 m C 2 ( m A 2 + m B 2 − m C 2 ) 2 (3.349) v A 2 = m A 4 + m B 4 + m C 4 − 2 m A 2 m B 2 − 2 m A 2 m C 2 − 2 m B 2 m C 2 m A 2 + m B 2 − m C 2 2 {:(3.349)v_(A)^(2)=(m_(A)^(4)+m_(B)^(4)+m_(C)^(4)-2m_(A)^(2)m_(B)^(2)-2m_(A)^(2)m_(C)^(2)-2m_(B)^(2)m_(C)^(2))/((m_(A)^(2)+m_(B)^(2)-m_(C)^(2))^(2)):}\begin{equation*} v_{A}^{2}=\frac{m_{A}^{4}+m_{B}^{4}+m_{C}^{4}-2 m_{A}^{2} m_{B}^{2}-2 m_{A}^{2} m_{C}^{2}-2 m_{B}^{2} m_{C}^{2}}{\left(m_{A}^{2}+m_{B}^{2}-m_{C}^{2}\right)^{2}} \tag{3.349} \end{equation*}(3.349)vA2=mA4+mB4+mC4−2mA2mB2−2mA2mC2−2mB2mC2(mA2+mB2−mC2)2

1.76

Conservation of 4 -momentum tells us that P = p 1 + p 2 P = p 1 + p 2 P=p_(1)+p_(2)P=p_{1}+p_{2}P=p1+p2, where P P PPP is the 4 -momentum of the new particle and p 1 p 1 p_(1)p_{1}p1 and p 2 p 2 p_(2)p_{2}p2 that of the initial two particles.
Squaring this relation we find that
(3.350) M 2 = P 2 = ( p 1 + p 2 ) 2 = m 1 2 + m 2 2 + 2 p 1 ⋅ p 2 (3.350) M 2 = P 2 = p 1 + p 2 2 = m 1 2 + m 2 2 + 2 p 1 ⋅ p 2 {:(3.350)M^(2)=P^(2)=(p_(1)+p_(2))^(2)=m_(1)^(2)+m_(2)^(2)+2p_(1)*p_(2):}\begin{equation*} M^{2}=P^{2}=\left(p_{1}+p_{2}\right)^{2}=m_{1}^{2}+m_{2}^{2}+2 p_{1} \cdot p_{2} \tag{3.350} \end{equation*}(3.350)M2=P2=(p1+p2)2=m12+m22+2p1⋅p2
In the rest frame of particle 2, we have p 1 = m γ ( 1 , v 1 ) p 1 = m γ 1 , v 1 p_(1)=m gamma(1,v_(1))p_{1}=m \gamma\left(1, \boldsymbol{v}_{1}\right)p1=mγ(1,v1) and p 2 = ( m , 0 ) p 2 = ( m , 0 ) p_(2)=(m,0)p_{2}=(m, \mathbf{0})p2=(m,0), leading to
(3.351) M = m 1 2 + m 2 2 + 2 m 1 m 2 1 − v 1 2 (3.351) M = m 1 2 + m 2 2 + 2 m 1 m 2 1 − v 1 2 {:(3.351)M=sqrt(m_(1)^(2)+m_(2)^(2)+(2m_(1)m_(2))/(sqrt(1-v_(1)^(2)))):}\begin{equation*} M=\sqrt{m_{1}^{2}+m_{2}^{2}+\frac{2 m_{1} m_{2}}{\sqrt{1-v_{1}^{2}}}} \tag{3.351} \end{equation*}(3.351)M=m12+m22+2m1m21−v12
In addition, using that v = p / E v = p / E v=p//E\boldsymbol{v}=\boldsymbol{p} / Ev=p/E for any particle and that the final 3-momentum is equal to the 3 -momentum of particle 1 in the rest frame of particle 2, we obtain
(3.352) v = p E = p 1 E = E 1 v 1 E 1 + m 2 = v 1 1 + m 2 / E 1 = m 1 m 1 + m 2 1 − v 1 2 v 1 (3.352) v = p E = p 1 E = E 1 v 1 E 1 + m 2 = v 1 1 + m 2 / E 1 = m 1 m 1 + m 2 1 − v 1 2 v 1 {:(3.352)v=(p)/(E)=(p_(1))/(E)=(E_(1)v_(1))/(E_(1)+m_(2))=(v_(1))/(1+m_(2)//E_(1))=(m_(1))/(m_(1)+m_(2)sqrt(1-v_(1)^(2)))v_(1):}\begin{equation*} \boldsymbol{v}=\frac{\boldsymbol{p}}{E}=\frac{\boldsymbol{p}_{1}}{E}=\frac{E_{1} \boldsymbol{v}_{1}}{E_{1}+m_{2}}=\frac{\boldsymbol{v}_{1}}{1+m_{2} / E_{1}}=\frac{m_{1}}{m_{1}+m_{2} \sqrt{1-v_{1}^{2}}} \boldsymbol{v}_{1} \tag{3.352} \end{equation*}(3.352)v=pE=p1E=E1v1E1+m2=v11+m2/E1=m1m1+m21−v12v1

1.77

a) Conservation of 4-momentum yields
(3.353) p = p 1 + p 2 ⇒ p 2 = ( p 1 + p 2 ) 2 = p 1 2 + p 2 2 + 2 p 1 ⋅ p 2 (3.353) p = p 1 + p 2 ⇒ p 2 = p 1 + p 2 2 = p 1 2 + p 2 2 + 2 p 1 ⋅ p 2 {:(3.353)p=p_(1)+p_(2)quad=>quadp^(2)=(p_(1)+p_(2))^(2)=p_(1)^(2)+p_(2)^(2)+2p_(1)*p_(2):}\begin{equation*} p=p_{1}+p_{2} \quad \Rightarrow \quad p^{2}=\left(p_{1}+p_{2}\right)^{2}=p_{1}^{2}+p_{2}^{2}+2 p_{1} \cdot p_{2} \tag{3.353} \end{equation*}(3.353)p=p1+p2⇒p2=(p1+p2)2=p12+p22+2p1⋅p2
which implies that
(3.354) m 2 = m 1 2 + m 2 2 + 2 ( m 1 2 + p 1 2 m 2 2 + p 2 2 − p 1 ⋅ p 2 ) (3.354) m 2 = m 1 2 + m 2 2 + 2 m 1 2 + p 1 2 m 2 2 + p 2 2 − p 1 ⋅ p 2 {:(3.354)m^(2)=m_(1)^(2)+m_(2)^(2)+2(sqrt(m_(1)^(2)+p_(1)^(2))sqrt(m_(2)^(2)+p_(2)^(2))-p_(1)*p_(2)):}\begin{equation*} m^{2}=m_{1}^{2}+m_{2}^{2}+2\left(\sqrt{m_{1}^{2}+\mathbf{p}_{1}^{2}} \sqrt{m_{2}^{2}+\mathbf{p}_{2}^{2}}-\mathbf{p}_{1} \cdot \mathbf{p}_{2}\right) \tag{3.354} \end{equation*}(3.354)m2=m12+m22+2(m12+p12m22+p22−p1⋅p2)
Assuming that the two particles move along the x x xxx-axis in the frame of some observer, we can express the 3-momenta of the two particles as
(3.355) p 1 = m 1 γ ( u 1 ) u 1 e x , p 2 = m 2 γ ( u 2 ) u 2 e x (3.355) p 1 = m 1 γ u 1 u 1 e x , p 2 = m 2 γ u 2 u 2 e x {:(3.355)p_(1)=m_(1)gamma(u_(1))u_(1)e_(x)","quadp_(2)=m_(2)gamma(u_(2))u_(2)e_(x):}\begin{equation*} \mathbf{p}_{1}=m_{1} \gamma\left(u_{1}\right) u_{1} \mathbf{e}_{x}, \quad \mathbf{p}_{2}=m_{2} \gamma\left(u_{2}\right) u_{2} \mathbf{e}_{x} \tag{3.355} \end{equation*}(3.355)p1=m1γ(u1)u1ex,p2=m2γ(u2)u2ex
where γ ( u i ) ≡ 1 / 1 − u i 2 , i = 1 , 2 γ u i ≡ 1 / 1 − u i 2 , i = 1 , 2 gamma(u_(i))-=1//sqrt(1-u_(i)^(2)),i=1,2\gamma\left(u_{i}\right) \equiv 1 / \sqrt{1-u_{i}^{2}}, i=1,2γ(ui)≡1/1−ui2,i=1,2. The energies of the two particles can then be rewritten as
m i 2 + p i 2 = m i 2 + m i 2 γ ( u i ) 2 u i 2 = m i 1 + γ ( u i ) 2 u i 2 = m i 1 + u i 2 1 − u i 2 (3.356) = m i 1 − u i 2 + u i 2 1 − u i 2 = m i 1 − u i 2 = m i γ ( u i ) m i 2 + p i 2 = m i 2 + m i 2 γ u i 2 u i 2 = m i 1 + γ u i 2 u i 2 = m i 1 + u i 2 1 − u i 2 (3.356) = m i 1 − u i 2 + u i 2 1 − u i 2 = m i 1 − u i 2 = m i γ u i {:[sqrt(m_(i)^(2)+p_(i)^(2))=sqrt(m_(i)^(2)+m_(i)^(2)gamma(u_(i))^(2)u_(i)^(2))=m_(i)sqrt(1+gamma(u_(i))^(2)u_(i)^(2))=m_(i)sqrt(1+(u_(i)^(2))/(1-u_(i)^(2)))],[(3.356)=m_(i)sqrt((1-u_(i)^(2)+u_(i)^(2))/(1-u_(i)^(2)))=(m_(i))/(sqrt(1-u_(i)^(2)))=m_(i)gamma(u_(i))]:}\begin{align*} \sqrt{m_{i}^{2}+\mathbf{p}_{i}^{2}} & =\sqrt{m_{i}^{2}+m_{i}^{2} \gamma\left(u_{i}\right)^{2} u_{i}^{2}}=m_{i} \sqrt{1+\gamma\left(u_{i}\right)^{2} u_{i}^{2}}=m_{i} \sqrt{1+\frac{u_{i}^{2}}{1-u_{i}^{2}}} \\ & =m_{i} \sqrt{\frac{1-u_{i}^{2}+u_{i}^{2}}{1-u_{i}^{2}}}=\frac{m_{i}}{\sqrt{1-u_{i}^{2}}}=m_{i} \gamma\left(u_{i}\right) \tag{3.356} \end{align*}mi2+pi2=mi2+mi2γ(ui)2ui2=mi1+γ(ui)2ui2=mi1+ui21−ui2(3.356)=mi1−ui2+ui21−ui2=mi1−ui2=miγ(ui)
Inserting the 3-momenta and the energies, we obtain
m 2 = m 1 2 + m 2 2 + 2 [ m 1 γ ( u 1 ) ⋅ m 2 γ ( u 2 ) − m 1 γ ( u 1 ) u 1 ⋅ m 2 γ ( u 2 ) u 2 ] (3.357) = m 1 2 + m 2 2 + 2 m 1 m 2 γ ( u 1 ) γ ( u 2 ) ( 1 − u 1 u 2 ) m 2 = m 1 2 + m 2 2 + 2 m 1 γ u 1 ⋅ m 2 γ u 2 − m 1 γ u 1 u 1 ⋅ m 2 γ u 2 u 2 (3.357) = m 1 2 + m 2 2 + 2 m 1 m 2 γ u 1 γ u 2 1 − u 1 u 2 {:[m^(2)=m_(1)^(2)+m_(2)^(2)+2[m_(1)gamma(u_(1))*m_(2)gamma(u_(2))-m_(1)gamma(u_(1))u_(1)*m_(2)gamma(u_(2))u_(2)]],[(3.357)=m_(1)^(2)+m_(2)^(2)+2m_(1)m_(2)gamma(u_(1))gamma(u_(2))(1-u_(1)u_(2))]:}\begin{align*} m^{2} & =m_{1}^{2}+m_{2}^{2}+2\left[m_{1} \gamma\left(u_{1}\right) \cdot m_{2} \gamma\left(u_{2}\right)-m_{1} \gamma\left(u_{1}\right) u_{1} \cdot m_{2} \gamma\left(u_{2}\right) u_{2}\right] \\ & =m_{1}^{2}+m_{2}^{2}+2 m_{1} m_{2} \gamma\left(u_{1}\right) \gamma\left(u_{2}\right)\left(1-u_{1} u_{2}\right) \tag{3.357} \end{align*}m2=m12+m22+2[m1γ(u1)⋅m2γ(u2)−m1γ(u1)u1⋅m2γ(u2)u2](3.357)=m12+m22+2m1m2γ(u1)γ(u2)(1−u1u2)
which is what we wanted to prove.
Now, we want to find u u uuu. Conservation of 3-momentum yields
(3.358) p = p 1 + p 2 (3.358) p = p 1 + p 2 {:(3.358)p=p_(1)+p_(2):}\begin{equation*} \mathbf{p}=\mathbf{p}_{1}+\mathbf{p}_{2} \tag{3.358} \end{equation*}(3.358)p=p1+p2
Assuming that the two particles move along the x x xxx-axis in the frame of the observer and using the expressions for the 3-momenta of the two particles, we find that
m γ ( u ) u = m 1 γ ( u 1 ) u 1 + m 2 γ ( u 2 ) u 2 (3.359) ⇒ [ m γ ( u ) u ] 2 = m 2 u 2 1 − u 2 = [ m 1 γ ( u 1 ) u 1 + m 2 γ ( u 2 ) u 2 ] 2 m γ ( u ) u = m 1 γ u 1 u 1 + m 2 γ u 2 u 2 (3.359) ⇒ [ m γ ( u ) u ] 2 = m 2 u 2 1 − u 2 = m 1 γ u 1 u 1 + m 2 γ u 2 u 2 2 {:[m gamma(u)u=m_(1)gamma(u_(1))u_(1)+m_(2)gamma(u_(2))u_(2)],[(3.359)quad=>quad[m gamma(u)u]^(2)=m^(2)(u^(2))/(1-u^(2))=[m_(1)gamma(u_(1))u_(1)+m_(2)gamma(u_(2))u_(2)]^(2)]:}\begin{align*} & m \gamma(u) u=m_{1} \gamma\left(u_{1}\right) u_{1}+m_{2} \gamma\left(u_{2}\right) u_{2} \\ & \quad \Rightarrow \quad[m \gamma(u) u]^{2}=m^{2} \frac{u^{2}}{1-u^{2}}=\left[m_{1} \gamma\left(u_{1}\right) u_{1}+m_{2} \gamma\left(u_{2}\right) u_{2}\right]^{2} \tag{3.359} \end{align*}mγ(u)u=m1γ(u1)u1+m2γ(u2)u2(3.359)⇒[mγ(u)u]2=m2u21−u2=[m1γ(u1)u1+m2γ(u2)u2]2
where we have used γ ( u ) ≡ 1 / 1 − u 2 γ ( u ) ≡ 1 / 1 − u 2 gamma(u)-=1//sqrt(1-u^(2))\gamma(u) \equiv 1 / \sqrt{1-u^{2}}γ(u)≡1/1−u2. This implies that
(3.360) u 2 1 − u 2 = [ m 1 γ ( u 1 ) u 1 + m 2 γ ( u 2 ) u 2 ] 2 m 2 ≡ M 2 (3.360) u 2 1 − u 2 = m 1 γ u 1 u 1 + m 2 γ u 2 u 2 2 m 2 ≡ M 2 {:(3.360)(u^(2))/(1-u^(2))=([m_(1)gamma(u_(1))u_(1)+m_(2)gamma(u_(2))u_(2)]^(2))/(m^(2))-=M^(2):}\begin{equation*} \frac{u^{2}}{1-u^{2}}=\frac{\left[m_{1} \gamma\left(u_{1}\right) u_{1}+m_{2} \gamma\left(u_{2}\right) u_{2}\right]^{2}}{m^{2}} \equiv M^{2} \tag{3.360} \end{equation*}(3.360)u21−u2=[m1γ(u1)u1+m2γ(u2)u2]2m2≡M2
Solving for u u uuu, we obtain
(3.361) u 2 = M 2 1 + M 2 ⇒ u = ± M 2 1 + M 2 (3.361) u 2 = M 2 1 + M 2 ⇒ u = ± M 2 1 + M 2 {:(3.361)u^(2)=(M^(2))/(1+M^(2))=>u=+-sqrt((M^(2))/(1+M^(2))):}\begin{equation*} u^{2}=\frac{M^{2}}{1+M^{2}} \Rightarrow u= \pm \sqrt{\frac{M^{2}}{1+M^{2}}} \tag{3.361} \end{equation*}(3.361)u2=M21+M2⇒u=±M21+M2
and reinserting M M MMM, we find the velocity
(3.362) u = ± [ m 1 γ ( u 1 ) u 1 + m 2 γ ( u 2 ) u 2 ] 2 m 2 1 + [ m 1 γ ( u 1 ) u 1 + m 2 γ ( u 2 ) u 2 ] 2 m 2 = ± m 1 γ ( u 1 ) u 1 + m 2 γ ( u 2 ) u 2 m 2 + [ m 1 γ ( u 1 ) u 1 + m 2 γ ( u 2 ) u 2 ] 2 . (3.362) u = ± m 1 γ u 1 u 1 + m 2 γ u 2 u 2 2 m 2 1 + m 1 γ u 1 u 1 + m 2 γ u 2 u 2 2 m 2 = ± m 1 γ u 1 u 1 + m 2 γ u 2 u 2 m 2 + m 1 γ u 1 u 1 + m 2 γ u 2 u 2 2 . {:(3.362)u=+-sqrt((([m_(1)gamma(u_(1))u_(1)+m_(2)gamma(u_(2))u_(2)]^(2))/(m^(2)))/(1+([m_(1)gamma(u_(1))u_(1)+m_(2)gamma(u_(2))u_(2)]^(2))/(m^(2))))=+-(m_(1)gamma(u_(1))u_(1)+m_(2)gamma(u_(2))u_(2))/(sqrt(m^(2)+[m_(1)gamma(u_(1))u_(1)+m_(2)gamma(u_(2))u_(2)]^(2))).:}\begin{equation*} u= \pm \sqrt{\frac{\frac{\left[m_{1} \gamma\left(u_{1}\right) u_{1}+m_{2} \gamma\left(u_{2}\right) u_{2}\right]^{2}}{m^{2}}}{1+\frac{\left[m_{1} \gamma\left(u_{1}\right) u_{1}+m_{2} \gamma\left(u_{2}\right) u_{2}\right]^{2}}{m^{2}}}}= \pm \frac{m_{1} \gamma\left(u_{1}\right) u_{1}+m_{2} \gamma\left(u_{2}\right) u_{2}}{\sqrt{m^{2}+\left[m_{1} \gamma\left(u_{1}\right) u_{1}+m_{2} \gamma\left(u_{2}\right) u_{2}\right]^{2}}} . \tag{3.362} \end{equation*}(3.362)u=±[m1γ(u1)u1+m2γ(u2)u2]2m21+[m1γ(u1)u1+m2γ(u2)u2]2m2=±m1γ(u1)u1+m2γ(u2)u2m2+[m1γ(u1)u1+m2γ(u2)u2]2.
Note that only the positive root is permissible.
b) In the rest frame of particle 1, we have
(3.363) u 1 = 0 , u 2 = v e x (3.363) u 1 = 0 , u 2 = v e x {:(3.363)u_(1)=0","quadu_(2)=ve_(x):}\begin{equation*} \mathbf{u}_{1}=\mathbf{0}, \quad \mathbf{u}_{2}=v \mathbf{e}_{x} \tag{3.363} \end{equation*}(3.363)u1=0,u2=vex
Inserting this into the expression for m m mmm from a), we find that
(3.364) m 2 = m 1 2 + m 2 2 + 2 m 1 m 2 γ ( 0 ) γ ( v ) ( 1 − 0 v ) = m 1 2 + m 2 2 + 2 m 1 m 2 γ ( v ) (3.364) m 2 = m 1 2 + m 2 2 + 2 m 1 m 2 γ ( 0 ) γ ( v ) ( 1 − 0 v ) = m 1 2 + m 2 2 + 2 m 1 m 2 γ ( v ) {:(3.364)m^(2)=m_(1)^(2)+m_(2)^(2)+2m_(1)m_(2)gamma(0)gamma(v)(1-0v)=m_(1)^(2)+m_(2)^(2)+2m_(1)m_(2)gamma(v):}\begin{equation*} m^{2}=m_{1}^{2}+m_{2}^{2}+2 m_{1} m_{2} \gamma(0) \gamma(v)(1-0 v)=m_{1}^{2}+m_{2}^{2}+2 m_{1} m_{2} \gamma(v) \tag{3.364} \end{equation*}(3.364)m2=m12+m22+2m1m2γ(0)γ(v)(1−0v)=m12+m22+2m1m2γ(v)
which is the required expression in terms of the relative velocity v v vvv.
c) In situation 1 ( u 1 = 0 ) 1 u 1 = 0 1(u_(1)=0)1\left(u_{1}=0\right)1(u1=0) we have u 2 = v u 2 = v u_(2)=vu_{2}=vu2=v. The total energy is therefore
(3.365) E tot , 1 = E 1 + E 2 = m 1 + m 2 1 − v 2 (3.365) E tot , 1 = E 1 + E 2 = m 1 + m 2 1 − v 2 {:(3.365)E_(tot,1)=E_(1)+E_(2)=m_(1)+(m_(2))/(sqrt(1-v^(2))):}\begin{equation*} E_{\mathrm{tot}, 1}=E_{1}+E_{2}=m_{1}+\frac{m_{2}}{\sqrt{1-v^{2}}} \tag{3.365} \end{equation*}(3.365)Etot,1=E1+E2=m1+m21−v2
In situation 2 ( m 1 γ ( u 1 ) u 1 = − m 2 γ ( u 2 ) u 2 ) 2 m 1 γ u 1 u 1 = − m 2 γ u 2 u 2 2(m_(1)gamma(u_(1))u_(1)=-m_(2)gamma(u_(2))u_(2))2\left(m_{1} \gamma\left(u_{1}\right) u_{1}=-m_{2} \gamma\left(u_{2}\right) u_{2}\right)2(m1γ(u1)u1=−m2γ(u2)u2), our frame is instead the rest frame of the new particle and the total energy is therefore
(3.366) E tot , 2 = m = m 1 2 + m 2 2 + 2 m 1 m 2 γ ( v ) (3.366) E tot , 2 = m = m 1 2 + m 2 2 + 2 m 1 m 2 γ ( v ) {:(3.366)E_(tot,2)=m=sqrt(m_(1)^(2)+m_(2)^(2)+2m_(1)m_(2)gamma(v)):}\begin{equation*} E_{\mathrm{tot}, 2}=m=\sqrt{m_{1}^{2}+m_{2}^{2}+2 m_{1} m_{2} \gamma(v)} \tag{3.366} \end{equation*}(3.366)Etot,2=m=m12+m22+2m1m2γ(v)
We find the difference between the energies to be
(3.367) Δ E = E tot , 1 − E tot , 2 = m 1 + m 2 1 − v 2 − m 1 2 + m 2 2 + 2 m 1 m 2 1 − v 2 (3.367) Δ E = E tot , 1 − E tot , 2 = m 1 + m 2 1 − v 2 − m 1 2 + m 2 2 + 2 m 1 m 2 1 − v 2 {:(3.367)Delta E=E_(tot,1)-E_(tot,2)=m_(1)+(m_(2))/(sqrt(1-v^(2)))-sqrt(m_(1)^(2)+m_(2)^(2)+2(m_(1)m_(2))/(sqrt(1-v^(2)))):}\begin{equation*} \Delta E=E_{\mathrm{tot}, 1}-E_{\mathrm{tot}, 2}=m_{1}+\frac{m_{2}}{\sqrt{1-v^{2}}}-\sqrt{m_{1}^{2}+m_{2}^{2}+2 \frac{m_{1} m_{2}}{\sqrt{1-v^{2}}}} \tag{3.367} \end{equation*}(3.367)ΔE=Etot,1−Etot,2=m1+m21−v2−m12+m22+2m1m21−v2
This may be rewritten as
(3.368) Δ E = m 1 + m 2 γ − ( m 1 + m 2 γ ) 2 − m 2 2 v 2 1 − v 2 ≥ 0 (3.368) Δ E = m 1 + m 2 γ − m 1 + m 2 γ 2 − m 2 2 v 2 1 − v 2 ≥ 0 {:(3.368)Delta E=m_(1)+m_(2)gamma-sqrt((m_(1)+m_(2)gamma)^(2)-(m_(2)^(2)v^(2))/(1-v^(2))) >= 0:}\begin{equation*} \Delta E=m_{1}+m_{2} \gamma-\sqrt{\left(m_{1}+m_{2} \gamma\right)^{2}-\frac{m_{2}^{2} v^{2}}{1-v^{2}}} \geq 0 \tag{3.368} \end{equation*}(3.368)ΔE=m1+m2γ−(m1+m2γ)2−m22v21−v2≥0
where γ = γ ( v ) γ = γ ( v ) gamma=gamma(v)\gamma=\gamma(v)γ=γ(v), with equality only if v = 0 v = 0 v=0v=0v=0. Therefore, more energy will be required in the rest frame of one of the particles than in the center-of-mass frame. For v ≪ 1 v ≪ 1 v≪1v \ll 1v≪1, keeping only terms up to second order in v v vvv, we find that
(3.369) Δ E ≃ ( m 2 − μ ) v 2 2 = m 2 2 v 2 2 ( m 1 + m 2 ) (3.369) Δ E ≃ m 2 − μ v 2 2 = m 2 2 v 2 2 m 1 + m 2 {:(3.369)Delta E≃(m_(2)-mu)(v^(2))/(2)=(m_(2)^(2)v^(2))/(2(m_(1)+m_(2))):}\begin{equation*} \Delta E \simeq\left(m_{2}-\mu\right) \frac{v^{2}}{2}=\frac{m_{2}^{2} v^{2}}{2\left(m_{1}+m_{2}\right)} \tag{3.369} \end{equation*}(3.369)ΔE≃(m2−μ)v22=m22v22(m1+m2)
where μ μ mu\muμ is the reduced mass μ = m 1 m 2 / ( m 1 + m 2 ) μ = m 1 m 2 / m 1 + m 2 mu=m_(1)m_(2)//(m_(1)+m_(2))\mu=m_{1} m_{2} /\left(m_{1}+m_{2}\right)μ=m1m2/(m1+m2) of the two-particle system.

1.78

Conservation of 4-momentum gives p π = p μ + p ν p π = p μ + p ν p_(pi)=p_(mu)+p_(nu)p_{\pi}=p_{\mu}+p_{\nu}pπ=pμ+pν, where p π = ( E π , p , 0 , 0 ) p π = E π , p , 0 , 0 p_(pi)=(E_(pi),p,0,0)p_{\pi}=\left(E_{\pi}, p, 0,0\right)pπ=(Eπ,p,0,0) and p μ = ( E μ , 0 , p ^ , 0 ) p μ = E μ , 0 , p ^ , 0 p_(mu)=(E_(mu),0,( hat(p)),0)p_{\mu}=\left(E_{\mu}, 0, \hat{p}, 0\right)pμ=(Eμ,0,p^,0). Here ( p , 0 , 0 ) ( p , 0 , 0 ) (p,0,0)(p, 0,0)(p,0,0) and ( 0 , p ^ , 0 ) ( 0 , p ^ , 0 ) (0, hat(p),0)(0, \hat{p}, 0)(0,p^,0) are the 3-momenta of the pion (in the x x xxx-direction) and the muon (in the y y yyy-direction), respectively, which are, however, not important for this problem. Taking the square of the 4-momentum relation after moving p μ p μ p_(mu)p_{\mu}pμ to the left-hand side, we find that
( p π − p μ ) 2 = p π 2 + p μ 2 − 2 p π ⋅ p μ = p v 2 = m v 2 = 0 (3.370) ⇒ m π 2 + m μ 2 − 2 E π E μ = 0 p π − p μ 2 = p π 2 + p μ 2 − 2 p π ⋅ p μ = p v 2 = m v 2 = 0 (3.370) ⇒ m π 2 + m μ 2 − 2 E π E μ = 0 {:[(p_(pi)-p_(mu))^(2)=p_(pi)^(2)+p_(mu)^(2)-2p_(pi)*p_(mu)=p_(v)^(2)=m_(v)^(2)=0],[(3.370)quad=>quadm_(pi)^(2)+m_(mu)^(2)-2E_(pi)E_(mu)=0]:}\begin{align*} & \left(p_{\pi}-p_{\mu}\right)^{2}=p_{\pi}^{2}+p_{\mu}^{2}-2 p_{\pi} \cdot p_{\mu}=p_{v}^{2}=m_{v}^{2}=0 \\ & \quad \Rightarrow \quad m_{\pi}^{2}+m_{\mu}^{2}-2 E_{\pi} E_{\mu}=0 \tag{3.370} \end{align*}(pπ−pμ)2=pπ2+pμ2−2pπ⋅pμ=pv2=mv2=0(3.370)⇒mπ2+mμ2−2EπEμ=0
Hence, we obtain the energy of the muon as
(3.371) E μ = m π 2 + m μ 2 2 E π = m π 2 + m μ 2 2 m π γ ( v ) (3.371) E μ = m π 2 + m μ 2 2 E π = m π 2 + m μ 2 2 m π γ ( v ) {:(3.371)E_(mu)=(m_(pi)^(2)+m_(mu)^(2))/(2E_(pi))=(m_(pi)^(2)+m_(mu)^(2))/(2m_(pi)gamma(v)):}\begin{equation*} E_{\mu}=\frac{m_{\pi}^{2}+m_{\mu}^{2}}{2 E_{\pi}}=\frac{m_{\pi}^{2}+m_{\mu}^{2}}{2 m_{\pi} \gamma(v)} \tag{3.371} \end{equation*}(3.371)Eμ=mπ2+mμ22Eπ=mπ2+mμ22mπγ(v)
where γ ( v ) ≡ 1 / 1 − v 2 γ ( v ) ≡ 1 / 1 − v 2 gamma(v)-=1//sqrt(1-v^(2))\gamma(v) \equiv 1 / \sqrt{1-v^{2}}γ(v)≡1/1−v2 and v v vvv is the velocity of the incoming pion.

1.79

Let the 4-momenta of the pion, the electron, and the antineutrino be p π , p e p π , p e p_(pi),p_(e)p_{\pi}, p_{e}pπ,pe, and p v p v p_(v)p_{v}pv, respectively. We find from energy-momentum conservation the relation
(3.372) m π 2 = p π 2 = ( p e + p v ) 2 (3.372) m π 2 = p π 2 = p e + p v 2 {:(3.372)m_(pi)^(2)=p_(pi)^(2)=(p_(e)+p_(v))^(2):}\begin{equation*} m_{\pi}^{2}=p_{\pi}^{2}=\left(p_{e}+p_{v}\right)^{2} \tag{3.372} \end{equation*}(3.372)mπ2=pπ2=(pe+pv)2
In the rest frame of the electron, we have p e = ( m , 0 ) p e = ( m , 0 ) p_(e)=(m,0)p_{e}=(m, \mathbf{0})pe=(m,0) and p v = ( E v , p ) p v = E v , p p_(v)=(E_(v),p)p_{v}=\left(E_{v}, \mathbf{p}\right)pv=(Ev,p), where m m mmm is the mass of the electron and E v = m v 2 + p 2 E v = m v 2 + p 2 E_(v)=sqrt(m_(v)^(2)+p^(2))E_{v}=\sqrt{m_{v}^{2}+\mathbf{p}^{2}}Ev=mv2+p2 and p p p\mathbf{p}p are the total energy and the
3-momentum of the antineutrino, respectively, m v m v m_(v)m_{v}mv being the mass of the antineutrino. Thus, we obtain
(3.373) m π 2 = ( p e + p v ) 2 = p e 2 + p v 2 + 2 p e ⋅ p v = m 2 + m v 2 + 2 m E v (3.373) m π 2 = p e + p v 2 = p e 2 + p v 2 + 2 p e ⋅ p v = m 2 + m v 2 + 2 m E v {:(3.373)m_(pi)^(2)=(p_(e)+p_(v))^(2)=p_(e)^(2)+p_(v)^(2)+2p_(e)*p_(v)=m^(2)+m_(v)^(2)+2mE_(v):}\begin{equation*} m_{\pi}^{2}=\left(p_{e}+p_{v}\right)^{2}=p_{e}^{2}+p_{v}^{2}+2 p_{e} \cdot p_{v}=m^{2}+m_{v}^{2}+2 m E_{v} \tag{3.373} \end{equation*}(3.373)mπ2=(pe+pv)2=pe2+pv2+2pe⋅pv=m2+mv2+2mEv
and hence, we have E v = Δ / ( 2 m ) E v = Δ / ( 2 m ) E_(v)=Delta//(2m)E_{v}=\Delta /(2 m)Ev=Δ/(2m), where Δ ≡ m π 2 − m 2 − m v 2 Δ ≡ m π 2 − m 2 − m v 2 Delta-=m_(pi)^(2)-m^(2)-m_(v)^(2)\Delta \equiv m_{\pi}^{2}-m^{2}-m_{v}^{2}Δ≡mπ2−m2−mv2. Using that the absolute value of the 3 -momentum of the antineutrino is
(3.374) | p | = E v 2 − m v 2 = Δ 2 4 m 2 − m v 2 (3.374) | p | = E v 2 − m v 2 = Δ 2 4 m 2 − m v 2 {:(3.374)|p|=sqrt(E_(v)^(2)-m_(v)^(2))=sqrt((Delta^(2))/(4m^(2))-m_(v)^(2)):}\begin{equation*} |\mathbf{p}|=\sqrt{E_{v}^{2}-m_{v}^{2}}=\sqrt{\frac{\Delta^{2}}{4 m^{2}}-m_{v}^{2}} \tag{3.374} \end{equation*}(3.374)|p|=Ev2−mv2=Δ24m2−mv2
we can calculate the velocity of the antineutrino as v = | p | / E v v = | p | / E v v=|p|//E_(v)v=|\mathbf{p}| / E_{v}v=|p|/Ev. The result is given by
(3.375) v = | p | E v = 1 − m v 2 E v 2 = 1 − 4 m 2 m v 2 Δ 2 = 1 − 4 m 2 m v 2 ( m π 2 − m 2 − m v 2 ) 2 (3.375) v = | p | E v = 1 − m v 2 E v 2 = 1 − 4 m 2 m v 2 Δ 2 = 1 − 4 m 2 m v 2 m π 2 − m 2 − m v 2 2 {:(3.375)v=(|p|)/(E_(v))=sqrt(1-(m_(v)^(2))/(E_(v)^(2)))=sqrt(1-(4m^(2)m_(v)^(2))/(Delta^(2)))=sqrt(1-(4m^(2)m_(v)^(2))/((m_(pi)^(2)-m^(2)-m_(v)^(2))^(2))):}\begin{equation*} v=\frac{|\mathbf{p}|}{E_{v}}=\sqrt{1-\frac{m_{v}^{2}}{E_{v}^{2}}}=\sqrt{1-\frac{4 m^{2} m_{v}^{2}}{\Delta^{2}}}=\sqrt{1-\frac{4 m^{2} m_{v}^{2}}{\left(m_{\pi}^{2}-m^{2}-m_{v}^{2}\right)^{2}}} \tag{3.375} \end{equation*}(3.375)v=|p|Ev=1−mv2Ev2=1−4m2mv2Δ2=1−4m2mv2(mπ2−m2−mv2)2
For the limiting value of v v vvv as the rest mass of the antineutrino goes to zero, since lim m v → 0 Δ = m π 2 − m 2 lim m v → 0   Δ = m π 2 − m 2 lim_(m_(v)rarr0)Delta=m_(pi)^(2)-m^(2)\lim _{m_{v} \rightarrow 0} \Delta=m_{\pi}^{2}-m^{2}limmv→0Δ=mπ2−m2, we find that v → 1 v → 1 v rarr1v \rightarrow 1v→1 as m v → 0 m v → 0 m_(v)rarr0m_{v} \rightarrow 0mv→0.

1.80

a) Consider the reaction π + ⟶ μ + + v μ π + ⟶ μ + + v μ pi^(+)longrightarrowmu^(+)+v_(mu)\pi^{+} \longrightarrow \mu^{+}+v_{\mu}π+⟶μ++vμ. Conservation of 4-momentum gives that
(3.376) P π + = P μ + + P v μ (3.376) P π + = P μ + + P v μ {:(3.376)P_(pi^(+))=P_(mu^(+))+P_(v_(mu)):}\begin{equation*} P_{\pi^{+}}=P_{\mu^{+}}+P_{v_{\mu}} \tag{3.376} \end{equation*}(3.376)Pπ+=Pμ++Pvμ
Subtracting P μ + P μ + P_(mu^(+))P_{\mu^{+}}Pμ+from both sides and squaring results in
(3.377) 0 = P v μ 2 = ( P π + − P μ + ) 2 = m π 2 + m μ 2 − 2 P π + ⋅ P μ + (3.377) 0 = P v μ 2 = P π + − P μ + 2 = m π 2 + m μ 2 − 2 P π + ⋅ P μ + {:(3.377)0=P_(v_(mu))^(2)=(P_(pi^(+))-P_(mu^(+)))^(2)=m_(pi)^(2)+m_(mu)^(2)-2P_(pi^(+))*P_(mu^(+)):}\begin{equation*} 0=P_{v_{\mu}}^{2}=\left(P_{\pi^{+}}-P_{\mu^{+}}\right)^{2}=m_{\pi}^{2}+m_{\mu}^{2}-2 P_{\pi^{+}} \cdot P_{\mu^{+}} \tag{3.377} \end{equation*}(3.377)0=Pvμ2=(Pπ+−Pμ+)2=mπ2+mμ2−2Pπ+⋅Pμ+
In the rest frame of the pion, we find that P π + ⋅ P μ + = m π E μ P π + ⋅ P μ + = m π E μ P_(pi^(+))*P_(mu^(+))=m_(pi)E_(mu)P_{\pi^{+}} \cdot P_{\mu^{+}}=m_{\pi} E_{\mu}Pπ+⋅Pμ+=mπEμ and therefore
(3.378) m π 2 + m μ 2 − 2 m π E μ = 0 (3.378) m π 2 + m μ 2 − 2 m π E μ = 0 {:(3.378)m_(pi)^(2)+m_(mu)^(2)-2m_(pi)E_(mu)=0:}\begin{equation*} m_{\pi}^{2}+m_{\mu}^{2}-2 m_{\pi} E_{\mu}=0 \tag{3.378} \end{equation*}(3.378)mπ2+mμ2−2mπEμ=0
Solving for the total energy E μ E μ E_(mu)E_{\mu}Eμ of the muon in the pion rest frame now results in
(3.379) E μ = m π 2 + m μ 2 2 m π (3.379) E μ = m π 2 + m μ 2 2 m π {:(3.379)E_(mu)=(m_(pi)^(2)+m_(mu)^(2))/(2m_(pi)):}\begin{equation*} E_{\mu}=\frac{m_{\pi}^{2}+m_{\mu}^{2}}{2 m_{\pi}} \tag{3.379} \end{equation*}(3.379)Eμ=mπ2+mμ22mπ
The kinetic energy of the muon is the difference between its total energy and its mass and we therefore find
(3.380) T μ + = E μ − m μ = m π 2 + m μ 2 2 m π − m μ = ( m π − m μ ) 2 2 m π (3.380) T μ + = E μ − m μ = m π 2 + m μ 2 2 m π − m μ = m π − m μ 2 2 m π {:(3.380)T_(mu^(+))=E_(mu)-m_(mu)=(m_(pi)^(2)+m_(mu)^(2))/(2m_(pi))-m_(mu)=((m_(pi)-m_(mu))^(2))/(2m_(pi)):}\begin{equation*} T_{\mu^{+}}=E_{\mu}-m_{\mu}=\frac{m_{\pi}^{2}+m_{\mu}^{2}}{2 m_{\pi}}-m_{\mu}=\frac{\left(m_{\pi}-m_{\mu}\right)^{2}}{2 m_{\pi}} \tag{3.380} \end{equation*}(3.380)Tμ+=Eμ−mμ=mπ2+mμ22mπ−mμ=(mπ−mμ)22mπ
Since the pion decays at rest, the absolute value of the neutrino momentum must equal that of the muon momentum. The energy-momentum relation for the pion therefore yields
(3.381) E μ 2 = m μ 2 + p v 2 ⟹ p v = m π 2 − m μ 2 2 m π (3.381) E μ 2 = m μ 2 + p v 2 ⟹ p v = m π 2 − m μ 2 2 m π {:(3.381)E_(mu)^(2)=m_(mu)^(2)+p_(v)^(2)quad Longrightarrowquadp_(v)=(m_(pi)^(2)-m_(mu)^(2))/(2m_(pi)):}\begin{equation*} E_{\mu}^{2}=m_{\mu}^{2}+p_{v}^{2} \quad \Longrightarrow \quad p_{v}=\frac{m_{\pi}^{2}-m_{\mu}^{2}}{2 m_{\pi}} \tag{3.381} \end{equation*}(3.381)Eμ2=mμ2+pv2⟹pv=mπ2−mμ22mπ
after inserting the expression for E μ E μ E_(mu)E_{\mu}Eμ and solving for p ν p ν p_(nu)p_{\nu}pν.
b) In the rest frame of the pion, the muon will travel the distance s ≡ γ ( v μ ) v μ τ μ = s ≡ γ v μ v μ τ μ = s-=gamma(v_(mu))v_(mu)tau_(mu)=s \equiv \gamma\left(v_{\mu}\right) v_{\mu} \tau_{\mu}=s≡γ(vμ)vμτμ= p μ τ μ / m μ p μ τ μ / m μ p_(mu)tau_(mu)//m_(mu)p_{\mu} \tau_{\mu} / m_{\mu}pμτμ/mμ before it decays. We therefore find that
(3.382) s = m π 2 − m μ 2 2 m π m μ τ μ (3.382) s = m π 2 − m μ 2 2 m π m μ τ μ {:(3.382)s=(m_(pi)^(2)-m_(mu)^(2))/(2m_(pi)m_(mu))tau_(mu):}\begin{equation*} s=\frac{m_{\pi}^{2}-m_{\mu}^{2}}{2 m_{\pi} m_{\mu}} \tau_{\mu} \tag{3.382} \end{equation*}(3.382)s=mπ2−mμ22mπmμτμ
since the muon and neutrino momenta are equal in magnitude.

1.81

In the rest frame of the decaying pion, the solution to Problem 1.80 resulted in
(3.383) E μ = m π 2 + m μ 2 2 m π and p = m π 2 − m μ 2 2 m π (3.383) E μ = m π 2 + m μ 2 2 m π  and  p = m π 2 − m μ 2 2 m π {:(3.383)E_(mu)=(m_(pi)^(2)+m_(mu)^(2))/(2m_(pi))quad" and "quad p=(m_(pi)^(2)-m_(mu)^(2))/(2m_(pi)):}\begin{equation*} E_{\mu}=\frac{m_{\pi}^{2}+m_{\mu}^{2}}{2 m_{\pi}} \quad \text { and } \quad p=\frac{m_{\pi}^{2}-m_{\mu}^{2}}{2 m_{\pi}} \tag{3.383} \end{equation*}(3.383)Eμ=mπ2+mμ22mπ and p=mπ2−mμ22mπ
for the muon energy and momentum, respectively. It follows that the 4-momentum of the muon in the rest frame of the pion is given by p μ = ( E μ , p ) p μ = E μ , p p_(mu)=(E_(mu),p)p_{\mu}=\left(E_{\mu}, p\right)pμ=(Eμ,p). However, we want to compute the energy of the muon in the rest frame of the Earth, and thus, we must Lorentz transform p μ p μ p_(mu)p_{\mu}pμ to this frame. The Lorentz transformation is in the opposite direction of the motion of the muon in the rest frame of the pion with velocity v = E π 2 − m π 2 / E π v = E π 2 − m π 2 / E π v=sqrt(E_(pi)^(2)-m_(pi)^(2))//E_(pi)v=\sqrt{E_{\pi}^{2}-m_{\pi}^{2}} / E_{\pi}v=Eπ2−mπ2/Eπ, where E π E π E_(pi)E_{\pi}Eπ is the energy of the pion in the rest frame of the Earth. It follows that in the rest frame of the Earth, we have
(3.384) E μ ′ = γ ( v ) ( E μ + v p ) = E π m π ( m π 2 + m μ 2 2 m π + m π 2 − m μ 2 2 m π 1 − m π 2 E π 2 ) (3.384) E μ ′ = γ ( v ) E μ + v p = E π m π m π 2 + m μ 2 2 m π + m π 2 − m μ 2 2 m π 1 − m π 2 E π 2 {:(3.384)E_(mu)^(')=gamma(v)(E_(mu)+vp)=(E_(pi))/(m_(pi))((m_(pi)^(2)+m_(mu)^(2))/(2m_(pi))+(m_(pi)^(2)-m_(mu)^(2))/(2m_(pi))sqrt(1-(m_(pi)^(2))/(E_(pi)^(2)))):}\begin{equation*} E_{\mu}^{\prime}=\gamma(v)\left(E_{\mu}+v p\right)=\frac{E_{\pi}}{m_{\pi}}\left(\frac{m_{\pi}^{2}+m_{\mu}^{2}}{2 m_{\pi}}+\frac{m_{\pi}^{2}-m_{\mu}^{2}}{2 m_{\pi}} \sqrt{1-\frac{m_{\pi}^{2}}{E_{\pi}^{2}}}\right) \tag{3.384} \end{equation*}(3.384)Eμ′=γ(v)(Eμ+vp)=Eπmπ(mπ2+mμ22mπ+mπ2−mμ22mπ1−mπ2Eπ2)
If we series expand the square root in the small quantity m π / E π m π / E π m_(pi)//E_(pi)m_{\pi} / E_{\pi}mπ/Eπ and keep the zeroth order term only, we obtain
(3.385) E μ ′ ≃ E π = 2 GeV (3.385) E μ ′ ≃ E π = 2 GeV {:(3.385)E_(mu)^(')≃E_(pi)=2GeV:}\begin{equation*} E_{\mu}^{\prime} \simeq E_{\pi}=2 \mathrm{GeV} \tag{3.385} \end{equation*}(3.385)Eμ′≃Eπ=2GeV

1.82

We study the decay R ⟶ μ + + μ − R ⟶ μ + + μ − R longrightarrowmu^(+)+mu^(-)R \longrightarrow \mu^{+}+\mu^{-}R⟶μ++μ−. In this decay, the total 4 -momentum must be preserved, and thus, we have
(3.386) p R = p μ + + p μ − (3.386) p R = p μ + + p μ − {:(3.386)p_(R)=p_(mu^(+))+p_(mu^(-)):}\begin{equation*} p_{R}=p_{\mu^{+}}+p_{\mu^{-}} \tag{3.386} \end{equation*}(3.386)pR=pμ++pμ−
The square of the mass M R M R M_(R)M_{R}MR of the resonance is given by the square of its 4-momentum, i.e.,
(3.387) M R 2 = p R 2 = p μ + 2 + p μ − 2 + 2 p μ + ⋅ p μ − = 2 ( m μ 2 + p μ + ⋅ p μ − ) (3.387) M R 2 = p R 2 = p μ + 2 + p μ − 2 + 2 p μ + ⋅ p μ − = 2 m μ 2 + p μ + ⋅ p μ − {:(3.387)M_(R)^(2)=p_(R)^(2)=p_(mu^(+))^(2)+p_(mu^(-))^(2)+2p_(mu^(+))*p_(mu^(-))=2(m_(mu)^(2)+p_(mu^(+))*p_(mu^(-))):}\begin{equation*} M_{R}^{2}=p_{R}^{2}=p_{\mu^{+}}^{2}+p_{\mu^{-}}^{2}+2 p_{\mu^{+}} \cdot p_{\mu^{-}}=2\left(m_{\mu}^{2}+p_{\mu^{+}} \cdot p_{\mu^{-}}\right) \tag{3.387} \end{equation*}(3.387)MR2=pR2=pμ+2+pμ−2+2pμ+⋅pμ−=2(mμ2+pμ+⋅pμ−)
If we place our coordinate system in such a way that the μ + μ + mu^(+)\mu^{+}μ+is traveling in the x x xxx-direction and the μ − μ − mu^(-)\mu^{-}μ−in the y y yyy-direction (this is possible, since the angle between the directions is 90 ∘ 90 ∘ 90^(@)90^{\circ}90∘ ), then the 4 -momentum of the muons will be
(3.388) p μ + = ( E , p , 0 , 0 ) and p μ − = ( E , 0 , p , 0 ) (3.388) p μ + = ( E , p , 0 , 0 )  and  p μ − = ( E , 0 , p , 0 ) {:(3.388)p_(mu^(+))=(E","p","0","0)quad" and "quadp_(mu^(-))=(E","0","p","0):}\begin{equation*} p_{\mu^{+}}=(E, p, 0,0) \quad \text { and } \quad p_{\mu^{-}}=(E, 0, p, 0) \tag{3.388} \end{equation*}(3.388)pμ+=(E,p,0,0) and pμ−=(E,0,p,0)
respectively, where p = 2.2 GeV p = 2.2 GeV p=2.2GeVp=2.2 \mathrm{GeV}p=2.2GeV and E = p 2 + m μ 2 E = p 2 + m μ 2 E=sqrt(p^(2)+m_(mu)^(2))E=\sqrt{p^{2}+m_{\mu}^{2}}E=p2+mμ2. It follows that
(3.389) p μ + ⋅ p μ − = E 2 = p 2 + m μ 2 (3.389) p μ + ⋅ p μ − = E 2 = p 2 + m μ 2 {:(3.389)p_(mu^(+))*p_(mu^(-))=E^(2)=p^(2)+m_(mu)^(2):}\begin{equation*} p_{\mu^{+}} \cdot p_{\mu^{-}}=E^{2}=p^{2}+m_{\mu}^{2} \tag{3.389} \end{equation*}(3.389)pμ+⋅pμ−=E2=p2+mμ2
and thus, we obtain
(3.390) M R 2 = 2 ( p 2 + 2 m μ 2 ) ≃ 2 p 2 ⇒ M R ≃ 2 p ≃ 3 GeV , (3.390) M R 2 = 2 p 2 + 2 m μ 2 ≃ 2 p 2 ⇒ M R ≃ 2 p ≃ 3 GeV , {:(3.390)M_(R)^(2)=2(p^(2)+2m_(mu)^(2))≃2p^(2)quad=>quadM_(R)≃sqrt2p≃3GeV",":}\begin{equation*} M_{R}^{2}=2\left(p^{2}+2 m_{\mu}^{2}\right) \simeq 2 p^{2} \quad \Rightarrow \quad M_{R} \simeq \sqrt{2} p \simeq 3 \mathrm{GeV}, \tag{3.390} \end{equation*}(3.390)MR2=2(p2+2mμ2)≃2p2⇒MR≃2p≃3GeV,
since p ≫ m μ p ≫ m μ p≫m_(mu)p \gg m_{\mu}p≫mμ.

1.83

There are several methods of solving this problem.
Method 1. We use the Doppler effect. In the system of the detector, the frequency of the photon is ω / h ω / h omega//h\omega / hω/h. This frequency is related to the frequency ω ′ / h ω ′ / h omega^(')//h\omega^{\prime} / hω′/h of the photon in the system of the decaying particle, from which it is emitted, according to the formula for the Doppler shift as
(3.391) ω h = ω ′ h 1 + v 1 − v (3.391) ω h = ω ′ h 1 + v 1 − v {:(3.391)(omega )/(h)=(omega^('))/(h)sqrt((1+v)/(1-v)):}\begin{equation*} \frac{\omega}{h}=\frac{\omega^{\prime}}{h} \sqrt{\frac{1+v}{1-v}} \tag{3.391} \end{equation*}(3.391)ωh=ω′h1+v1−v
where v v vvv is the velocity of the decaying particle before it emits the photon. The frequency must be blueshifted, since the emitting particle moves toward the detector. Now, we have v = | p | / E v = | p | / E v=|p|//Ev=|\boldsymbol{p}| / Ev=|p|/E. Inserting this into the Doppler formula above gives, after some simplifications,
(3.392) ω h = ω ′ h E + | p | M (3.392) ω h = ω ′ h E + | p | M {:(3.392)(omega )/(h)=(omega^('))/(h)(E+|p|)/(M):}\begin{equation*} \frac{\omega}{h}=\frac{\omega^{\prime}}{h} \frac{E+|\boldsymbol{p}|}{M} \tag{3.392} \end{equation*}(3.392)ωh=ω′hE+|p|M
since E 2 − p 2 = M 2 E 2 − p 2 = M 2 E^(2)-p^(2)=M^(2)E^{2}-\boldsymbol{p}^{2}=M^{2}E2−p2=M2. Thus, solving for ω ′ ω ′ omega^(')\omega^{\prime}ω′ gives the answer
(3.393) ω ′ = ω M E + | p | (3.393) ω ′ = ω M E + | p | {:(3.393)omega^(')=omega(M)/(E+|p|):}\begin{equation*} \omega^{\prime}=\omega \frac{M}{E+|\boldsymbol{p}|} \tag{3.393} \end{equation*}(3.393)ω′=ωME+|p|
Method 2. Conservation of energy and momentum says that if the decaying particle emits a photon with 4-momentum k = ( w , k ) k = ( w , k ) k=(w,k)k=(w, \boldsymbol{k})k=(w,k) and a rest product (which can be several particles) of momentum p ′ = ( E ′ , p ′ ) p ′ = E ′ , p ′ p^(')=(E^('),p^('))p^{\prime}=\left(E^{\prime}, \boldsymbol{p}^{\prime}\right)p′=(E′,p′), then we have, in the rest frame of the decaying particle,
(3.394) M = E ′ + ω ′ (3.394) M = E ′ + ω ′ {:(3.394)M=E^(')+omega^('):}\begin{equation*} M=E^{\prime}+\omega^{\prime} \tag{3.394} \end{equation*}(3.394)M=E′+ω′
In this system, conservation of momentum reads p ′ = − k p ′ = − k p^(')=-k\boldsymbol{p}^{\prime}=-\boldsymbol{k}p′=−k. We then obtain
(3.395) ( M − ω ′ ) 2 = E ′ 2 = p ′ 2 + ( p ′ ) 2 = p ′ 2 + ω ′ 2 (3.395) M − ω ′ 2 = E ′ 2 = p ′ 2 + p ′ 2 = p ′ 2 + ω ′ 2 {:(3.395)(M-omega^('))^(2)=E^('2)=p^('2)+(p^('))^(2)=p^('2)+omega^('2):}\begin{equation*} \left(M-\omega^{\prime}\right)^{2}=E^{\prime 2}=p^{\prime 2}+\left(p^{\prime}\right)^{2}=p^{\prime 2}+\omega^{\prime 2} \tag{3.395} \end{equation*}(3.395)(M−ω′)2=E′2=p′2+(p′)2=p′2+ω′2
where p ′ 2 p ′ 2 p^('2)p^{\prime 2}p′2 is the 4-momentum squared of p ′ p ′ p^(')p^{\prime}p′. However, conservation of 4-momentum also gives
(3.396) p = p ′ + k (3.396) p = p ′ + k {:(3.396)p=p^(')+k:}\begin{equation*} p=p^{\prime}+k \tag{3.396} \end{equation*}(3.396)p=p′+k
whence p ′ 2 = ( p − k ) 2 = M 2 − 2 p ⋅ k p ′ 2 = ( p − k ) 2 = M 2 − 2 p ⋅ k p^('2)=(p-k)^(2)=M^(2)-2p*kp^{\prime 2}=(p-k)^{2}=M^{2}-2 p \cdot kp′2=(p−k)2=M2−2p⋅k, where we have used k 2 = 0 k 2 = 0 k^(2)=0k^{2}=0k2=0. Inserting this into the first relation gives
(3.397) ( M − ω ′ ) 2 = M 2 − 2 p ⋅ k + ω ′ 2 (3.397) M − ω ′ 2 = M 2 − 2 p ⋅ k + ω ′ 2 {:(3.397)(M-omega^('))^(2)=M^(2)-2p*k+omega^('2):}\begin{equation*} \left(M-\omega^{\prime}\right)^{2}=M^{2}-2 p \cdot k+\omega^{\prime 2} \tag{3.397} \end{equation*}(3.397)(M−ω′)2=M2−2p⋅k+ω′2
If we solve for ω ′ ω ′ omega^(')\omega^{\prime}ω′, we obtain ω ′ = p ⋅ k / M ω ′ = p ⋅ k / M omega^(')=p*k//M\omega^{\prime}=p \cdot k / Mω′=p⋅k/M. By inserting p = ( E , p ) p = ( E , p ) p=(E,p)p=(E, \boldsymbol{p})p=(E,p) and k = ( ω , k ) k = ( ω , k ) k=(omega,k)k=(\omega, \boldsymbol{k})k=(ω,k) with p p ppp parallel to k k kkk, we obtain the same answer as with Method 1 .
Method 3. The previous method suggests that one can solve the problem by studying the relativistic invariant p ⋅ k p ⋅ k p*kp \cdot kp⋅k. In the rest frame of the decaying particle, its value is M ω ′ M ω ′ Momega^(')M \omega^{\prime}Mω′. In the frame of the detector, its value is
(3.398) p ⋅ k = E ω − p ⋅ k = ω ( E − | p | ) (3.398) p ⋅ k = E ω − p ⋅ k = ω ( E − | p | ) {:(3.398)p*k=E omega-p*k=omega(E-|p|):}\begin{equation*} p \cdot k=E \omega-\boldsymbol{p} \cdot \boldsymbol{k}=\omega(E-|\boldsymbol{p}|) \tag{3.398} \end{equation*}(3.398)p⋅k=Eω−p⋅k=ω(E−|p|)
Since the value of the invariant is independent of the frame, one finds M ω ′ = M ω ′ = Momega^(')=M \omega^{\prime}=Mω′= ω ( E − | p | ) ω ( E − | p | ) omega(E-|p|)\omega(E-|\boldsymbol{p}|)ω(E−|p|), which, after simplifications, leads to the same answer as with the previous two methods.
Method 4. Lastly, one can also simply make a Lorentz transformation of k = k = k=k=k= ( ω ′ , ω ′ , 0 , 0 ) ω ′ , ω ′ , 0 , 0 (omega^('),omega^('),0,0)\left(\omega^{\prime}, \omega^{\prime}, 0,0\right)(ω′,ω′,0,0), which is 4 -wavevector of the photon in the rest frame of the decaying particle and where we have put the direction to the detector to coincide with the x x xxx-axis to the detector system in which the particle moves with speed v = | p | / E v = | p | / E v=|p|//Ev=|\boldsymbol{p}| / Ev=|p|/E toward the detector. The detector then moves with speed − v − v -v-v−v relative to the particle. For the 0 -component, one then finds that
(3.399) ω = ω ′ γ + ω ′ γ v = ω ′ γ ( 1 + v ) (3.399) ω = ω ′ γ + ω ′ γ v = ω ′ γ ( 1 + v ) {:(3.399)omega=omega^(')gamma+omega^(')gamma v=omega^(')gamma(1+v):}\begin{equation*} \omega=\omega^{\prime} \gamma+\omega^{\prime} \gamma v=\omega^{\prime} \gamma(1+v) \tag{3.399} \end{equation*}(3.399)ω=ω′γ+ω′γv=ω′γ(1+v)
where γ ≡ 1 / 1 − v 2 γ ≡ 1 / 1 − v 2 gamma-=1//sqrt(1-v^(2))\gamma \equiv 1 / \sqrt{1-v^{2}}γ≡1/1−v2. Inserting the value of v v vvv above gives, after simplifications, the same result as obtained by the other three methods.

1.84

In the rest frame of the positron, let the 4-momenta be p e = ( E e / c , 0 , 0 , p ) p e = E e / c , 0 , 0 , p p_(e)=(E_(e)//c,0,0,p)p_{e}=\left(E_{e} / c, 0,0, p\right)pe=(Ee/c,0,0,p) and p p = ( m e c , 0 , 0 , 0 ) p p = m e c , 0 , 0 , 0 p_(p)=(m_(e)c,0,0,0)p_{p}=\left(m_{e} c, 0,0,0\right)pp=(mec,0,0,0) for the electron and the positron, respectively, where E e E e E_(e)E_{e}Ee is the total energy of the electron, ( 0 , 0 , p ) ( 0 , 0 , p ) (0,0,p)(0,0, p)(0,0,p) is the 3 -momentum of the electron, and m e m e m_(e)m_{e}me is the mass of an electron or a positron. The 4-momenta of the photons are
k 1 = ( ω 1 , ω 1 sin ⁡ ϕ , 0 , ω 1 cos ⁡ ϕ ) k 1 = ω 1 , ω 1 sin ⁡ ϕ , 0 , ω 1 cos ⁡ ϕ k_(1)=(omega_(1),omega_(1)sin phi,0,omega_(1)cos phi)k_{1}=\left(\omega_{1}, \omega_{1} \sin \phi, 0, \omega_{1} \cos \phi\right)k1=(ω1,ω1sin⁡ϕ,0,ω1cos⁡ϕ) and k 2 = ( ω 2 , − ω 2 sin ⁡ ϕ , 0 , ω 2 cos ⁡ ϕ ) k 2 = ω 2 , − ω 2 sin ⁡ ϕ , 0 , ω 2 cos ⁡ ϕ k_(2)=(omega_(2),-omega_(2)sin phi,0,omega_(2)cos phi)k_{2}=\left(\omega_{2},-\omega_{2} \sin \phi, 0, \omega_{2} \cos \phi\right)k2=(ω2,−ω2sin⁡ϕ,0,ω2cos⁡ϕ), respectively. Conservation of 4-momentum gives
(3.400) p e + p p = k 1 + k 2 , (3.400) p e + p p = k 1 + k 2 , {:(3.400)p_(e)+p_(p)=k_(1)+k_(2)",":}\begin{equation*} p_{e}+p_{p}=k_{1}+k_{2}, \tag{3.400} \end{equation*}(3.400)pe+pp=k1+k2,
i.e., in the rest frame of the positron, we have
(3.401) E e c + m e c = ω 1 + ω 2 (3.402) 0 = ω 1 sin ⁡ ϕ − ω 2 sin ⁡ ϕ (3.403) p = ω 1 cos ⁡ ϕ + ω 2 cos ⁡ ϕ (3.401) E e c + m e c = ω 1 + ω 2 (3.402) 0 = ω 1 sin ⁡ ϕ − ω 2 sin ⁡ ϕ (3.403) p = ω 1 cos ⁡ ϕ + ω 2 cos ⁡ ϕ {:[(3.401)(E_(e))/(c)+m_(e)c=omega_(1)+omega_(2)],[(3.402)0=omega_(1)sin phi-omega_(2)sin phi],[(3.403)p=omega_(1)cos phi+omega_(2)cos phi]:}\begin{align*} \frac{E_{e}}{c}+m_{e} c & =\omega_{1}+\omega_{2} \tag{3.401}\\ 0 & =\omega_{1} \sin \phi-\omega_{2} \sin \phi \tag{3.402}\\ p & =\omega_{1} \cos \phi+\omega_{2} \cos \phi \tag{3.403} \end{align*}(3.401)Eec+mec=ω1+ω2(3.402)0=ω1sin⁡ϕ−ω2sin⁡ϕ(3.403)p=ω1cos⁡ϕ+ω2cos⁡ϕ
Hence, we find that ω 1 = ω 2 ω 1 = ω 2 omega_(1)=omega_(2)\omega_{1}=\omega_{2}ω1=ω2.
a) From the relations above, we obtain the angle ϕ ϕ phi\phiϕ as a function of the total energy of the electron E e E e E_(e)E_{e}Ee as
cos ⁡ ϕ = p k 1 + k 2 = p ω 1 + ω 2 = p E e c + m e c = E e 2 c 2 − m e 2 c 2 E e c + m e c = E e − m e c 2 E e + m e c 2 (3.404) ⇒ ϕ = arccos ⁡ E e − m e c 2 E e + m e c 2 cos ⁡ ϕ = p k 1 + k 2 = p ω 1 + ω 2 = p E e c + m e c = E e 2 c 2 − m e 2 c 2 E e c + m e c = E e − m e c 2 E e + m e c 2 (3.404) ⇒ ϕ = arccos ⁡ E e − m e c 2 E e + m e c 2 {:[cos phi=(p)/(k_(1)+k_(2))=(p)/(omega_(1)+omega_(2))=(p)/((E_(e))/(c)+m_(e)c)=(sqrt((E_(e)^(2))/(c^(2))-m_(e)^(2)c^(2)))/((E_(e))/(c)+m_(e)c)=sqrt((E_(e)-m_(e)c^(2))/(E_(e)+m_(e)c^(2)))],[(3.404)quad=>quad phi=arccos sqrt((E_(e)-m_(e)c^(2))/(E_(e)+m_(e)c^(2)))]:}\begin{align*} & \cos \phi=\frac{p}{k_{1}+k_{2}}=\frac{p}{\omega_{1}+\omega_{2}}=\frac{p}{\frac{E_{e}}{c}+m_{e} c}=\frac{\sqrt{\frac{E_{e}^{2}}{c^{2}}-m_{e}^{2} c^{2}}}{\frac{E_{e}}{c}+m_{e} c}=\sqrt{\frac{E_{e}-m_{e} c^{2}}{E_{e}+m_{e} c^{2}}} \\ & \quad \Rightarrow \quad \phi=\arccos \sqrt{\frac{E_{e}-m_{e} c^{2}}{E_{e}+m_{e} c^{2}}} \tag{3.404} \end{align*}cos⁡ϕ=pk1+k2=pω1+ω2=pEec+mec=Ee2c2−me2c2Eec+mec=Ee−mec2Ee+mec2(3.404)⇒ϕ=arccos⁡Ee−mec2Ee+mec2
b) In the nonrelativistic limit, i.e., E e ≃ m e c 2 + p 2 / ( 2 m e ) E e ≃ m e c 2 + p 2 / 2 m e E_(e)≃m_(e)c^(2)+p^(2)//(2m_(e))E_{e} \simeq m_{e} c^{2}+p^{2} /\left(2 m_{e}\right)Ee≃mec2+p2/(2me), with p ≪ m 2 p ≪ m 2 p≪m_(2)p \ll m_{2}p≪m2, we find that
(3.405) cos ⁡ ϕ ≃ p 2 4 m e 2 c 2 + p 2 ≃ p 2 m e c = { p ≃ m e v } = v 2 c . (3.405) cos ⁡ ϕ ≃ p 2 4 m e 2 c 2 + p 2 ≃ p 2 m e c = p ≃ m e v = v 2 c . {:(3.405)cos phi≃sqrt((p^(2))/(4m_(e)^(2)c^(2)+p^(2)))≃(p)/(2m_(e)c)={p≃m_(e)v}=(v)/(2c).:}\begin{equation*} \cos \phi \simeq \sqrt{\frac{p^{2}}{4 m_{e}^{2} c^{2}+p^{2}}} \simeq \frac{p}{2 m_{e} c}=\left\{p \simeq m_{e} v\right\}=\frac{v}{2 c} . \tag{3.405} \end{equation*}(3.405)cos⁡ϕ≃p24me2c2+p2≃p2mec={p≃mev}=v2c.

1.85

Conservation of 4-momentum gives
(3.406) P 1 + P 2 = P + P ′ (3.406) P 1 + P 2 = P + P ′ {:(3.406)P_(1)+P_(2)=P+P^('):}\begin{equation*} P_{1}+P_{2}=P+P^{\prime} \tag{3.406} \end{equation*}(3.406)P1+P2=P+P′
which implies that P ′ = P 1 + P 2 − P P ′ = P 1 + P 2 − P P^(')=P_(1)+P_(2)-PP^{\prime}=P_{1}+P_{2}-PP′=P1+P2−P. Now, photons are lightlike. Thus, we find that
(3.407) P ′ 2 = 0 = ( P 1 + P 2 − P ) 2 = 2 ( P 1 ⋅ P 2 − P 1 ⋅ P − P 2 ⋅ P ) (3.407) P ′ 2 = 0 = P 1 + P 2 − P 2 = 2 P 1 ⋅ P 2 − P 1 ⋅ P − P 2 ⋅ P {:(3.407)P^('2)=0=(P_(1)+P_(2)-P)^(2)=2(P_(1)*P_(2)-P_(1)*P-P_(2)*P):}\begin{equation*} P^{\prime 2}=0=\left(P_{1}+P_{2}-P\right)^{2}=2\left(P_{1} \cdot P_{2}-P_{1} \cdot P-P_{2} \cdot P\right) \tag{3.407} \end{equation*}(3.407)P′2=0=(P1+P2−P)2=2(P1⋅P2−P1⋅P−P2⋅P)
since P 1 2 = P 2 2 = P 2 = 0 P 1 2 = P 2 2 = P 2 = 0 P_(1)^(2)=P_(2)^(2)=P^(2)=0P_{1}{ }^{2}=P_{2}^{2}=P^{2}=0P12=P22=P2=0. Therefore, we have
(3.408) P 1 ⋅ P 2 − P 1 ⋅ P − P 2 ⋅ P = 0 (3.408) P 1 ⋅ P 2 − P 1 ⋅ P − P 2 ⋅ P = 0 {:(3.408)P_(1)*P_(2)-P_(1)*P-P_(2)*P=0:}\begin{equation*} P_{1} \cdot P_{2}-P_{1} \cdot P-P_{2} \cdot P=0 \tag{3.408} \end{equation*}(3.408)P1⋅P2−P1⋅P−P2⋅P=0
The energy for a photon is E = p c = { p = h λ } = h c λ E = p c = p = h λ = h c λ E=pc={p=(h)/( lambda)}=(hc)/(lambda)E=p c=\left\{p=\frac{h}{\lambda}\right\}=\frac{h c}{\lambda}E=pc={p=hλ}=hcλ. Using P 1 = h λ 1 ( 1 , 1 , 0 ) P 1 = h λ 1 ( 1 , 1 , 0 ) P_(1)=(h)/(lambda_(1))(1,1,0)P_{1}=\frac{h}{\lambda_{1}}(1,1,0)P1=hλ1(1,1,0), P 2 = h λ 2 ( 1 , − 1 , 0 ) P 2 = h λ 2 ( 1 , − 1 , 0 ) P_(2)=(h)/(lambda_(2))(1,-1,0)P_{2}=\frac{h}{\lambda_{2}}(1,-1,0)P2=hλ2(1,−1,0), and P = h λ ( 1 , cos ⁡ θ , sin ⁡ θ ) P = h λ ( 1 , cos ⁡ θ , sin ⁡ θ ) P=(h)/( lambda)(1,cos theta,sin theta)P=\frac{h}{\lambda}(1, \cos \theta, \sin \theta)P=hλ(1,cos⁡θ,sin⁡θ), we obtain
(3.409) 2 h 2 λ 1 λ 2 = P 1 ⋅ P 2 = P ⋅ ( P 1 + P 2 ) = h 2 λ ⋅ [ 1 − cos ⁡ θ λ 1 + 1 + cos ⁡ θ λ 2 ] (3.409) 2 h 2 λ 1 λ 2 = P 1 ⋅ P 2 = P ⋅ P 1 + P 2 = h 2 λ ⋅ 1 − cos ⁡ θ λ 1 + 1 + cos ⁡ θ λ 2 {:(3.409)(2h^(2))/(lambda_(1)lambda_(2))=P_(1)*P_(2)=P*(P_(1)+P_(2))=(h^(2))/(lambda)*[(1-cos theta)/(lambda_(1))+(1+cos theta)/(lambda_(2))]:}\begin{equation*} \frac{2 h^{2}}{\lambda_{1} \lambda_{2}}=P_{1} \cdot P_{2}=P \cdot\left(P_{1}+P_{2}\right)=\frac{h^{2}}{\lambda} \cdot\left[\frac{1-\cos \theta}{\lambda_{1}}+\frac{1+\cos \theta}{\lambda_{2}}\right] \tag{3.409} \end{equation*}(3.409)2h2λ1λ2=P1⋅P2=P⋅(P1+P2)=h2λ⋅[1−cos⁡θλ1+1+cos⁡θλ2]
Solving for λ λ lambda\lambdaλ now results in
(3.410) λ = 1 2 [ λ 2 ( 1 − cos ⁡ θ ) + λ 1 ( 1 + cos ⁡ θ ) ] . (3.410) λ = 1 2 λ 2 ( 1 − cos ⁡ θ ) + λ 1 ( 1 + cos ⁡ θ ) . {:(3.410)lambda=(1)/(2)[lambda_(2)(1-cos theta)+lambda_(1)(1+cos theta)].:}\begin{equation*} \lambda=\frac{1}{2}\left[\lambda_{2}(1-\cos \theta)+\lambda_{1}(1+\cos \theta)\right] . \tag{3.410} \end{equation*}(3.410)λ=12[λ2(1−cos⁡θ)+λ1(1+cos⁡θ)].

1.86

First, define the Lorentz invariant total 4 -momentum squared of the pions as s = ( p 1 + p 2 ) 2 s = p 1 + p 2 2 s=(p_(1)+p_(2))^(2)s=\left(p_{1}+p_{2}\right)^{2}s=(p1+p2)2. Using conservation of 4-momentum, i.e., P = p 1 + p 2 + k P = p 1 + p 2 + k P=p_(1)+p_(2)+kP=p_{1}+p_{2}+kP=p1+p2+k, we can then calculate s s sss to be s = ( P − k ) 2 = M ( M − 2 ω ) s = ( P − k ) 2 = M ( M − 2 ω ) s=(P-k)^(2)=M(M-2omega)s=(P-k)^{2}=M(M-2 \omega)s=(P−k)2=M(M−2ω). However, in the rest frame of the pions, we know that s = 2 E s = 2 E sqrts=2E\sqrt{s}=2 Es=2E, where E = m 2 + p 2 E = m 2 + p 2 E=sqrt(m^(2)+p^(2))E=\sqrt{m^{2}+p^{2}}E=m2+p2 and p 2 = p 1 2 = p 2 2 p 2 = p 1 2 = p 2 2 p^(2)=p_(1)^(2)=p_(2)^(2)p^{2}=p_{1}^{2}=p_{2}^{2}p2=p12=p22. Therefore, using the two expressions for s s sss, we obtain M 2 − 2 M ω = 4 ( m 2 + p 2 ) M 2 − 2 M ω = 4 m 2 + p 2 M^(2)-2M omega=4(m^(2)+p^(2))M^{2}-2 M \omega=4\left(m^{2}+p^{2}\right)M2−2Mω=4(m2+p2), which can be solved for p p ppp to give
(3.411) p = 1 2 M 2 − 2 M ω − 4 m 2 (3.411) p = 1 2 M 2 − 2 M ω − 4 m 2 {:(3.411)p=(1)/(2)sqrt(M^(2)-2M omega-4m^(2)):}\begin{equation*} p=\frac{1}{2} \sqrt{M^{2}-2 M \omega-4 m^{2}} \tag{3.411} \end{equation*}(3.411)p=12M2−2Mω−4m2
Finally, the speed of the pions relative to their center-of-mass frame is v = p / E v = p / E v=p//Ev=p / Ev=p/E, so inserting the expression for p p ppp and E = 1 2 s = 1 2 M 2 − 2 M ω E = 1 2 s = 1 2 M 2 − 2 M ω E=(1)/(2)sqrts=(1)/(2)sqrt(M^(2)-2M omega)E=\frac{1}{2} \sqrt{s}=\frac{1}{2} \sqrt{M^{2}-2 M \omega}E=12s=12M2−2Mω, we find that
(3.412) v = p E = M 2 − 2 M ω − 4 m 2 M 2 − 2 M ω = 1 − 4 m 2 M ( M − 2 ω ) . (3.412) v = p E = M 2 − 2 M ω − 4 m 2 M 2 − 2 M ω = 1 − 4 m 2 M ( M − 2 ω ) . {:(3.412)v=(p)/(E)=sqrt((M^(2)-2M omega-4m^(2))/(M^(2)-2M omega))=sqrt(1-(4m^(2))/(M(M-2omega))).:}\begin{equation*} v=\frac{p}{E}=\sqrt{\frac{M^{2}-2 M \omega-4 m^{2}}{M^{2}-2 M \omega}}=\sqrt{1-\frac{4 m^{2}}{M(M-2 \omega)}} . \tag{3.412} \end{equation*}(3.412)v=pE=M2−2Mω−4m2M2−2Mω=1−4m2M(M−2ω).
1.87
Before the decay, the Σ 0 Σ 0 Sigma^(0)\Sigma^{0}Σ0 particle moves with 4-momentum ( E , p E , p E,pE, \boldsymbol{p}E,p ) toward the detector. After the decay, the Λ Λ Lambda\LambdaΛ particle moves toward the detector with 4-momentum ( E ′ , p ′ E ′ , p ′ E^('),p^(')E^{\prime}, \boldsymbol{p}^{\prime}E′,p′ ) and the photon with momentum ( ω ′ , k ′ ω ′ , k ′ omega^('),k^(')\omega^{\prime}, \boldsymbol{k}^{\prime}ω′,k′ ).
a) The total energy of the Σ 0 Σ 0 Sigma^(0)\Sigma^{0}Σ0 particle is given by
(3.413) E Σ = m Σ 1 − v 2 = m Σ 1 − 1 / 3 2 = 3 m Σ 2 2 (3.413) E Σ = m Σ 1 − v 2 = m Σ 1 − 1 / 3 2 = 3 m Σ 2 2 {:(3.413)E_(Sigma)=(m_(Sigma))/(sqrt(1-v^(2)))=(m_(Sigma))/(sqrt(1-1//3^(2)))=(3m_(Sigma))/(2sqrt2):}\begin{equation*} E_{\Sigma}=\frac{m_{\Sigma}}{\sqrt{1-v^{2}}}=\frac{m_{\Sigma}}{\sqrt{1-1 / 3^{2}}}=\frac{3 m_{\Sigma}}{2 \sqrt{2}} \tag{3.413} \end{equation*}(3.413)EΣ=mΣ1−v2=mΣ1−1/32=3mΣ22
b) Conservation of 4-momentum gives the relation
(3.414) P Σ = P γ + P Λ (3.414) P Σ = P γ + P Λ {:(3.414)P_(Sigma)=P_(gamma)+P_(Lambda):}\begin{equation*} P_{\Sigma}=P_{\gamma}+P_{\Lambda} \tag{3.414} \end{equation*}(3.414)PΣ=Pγ+PΛ
Subtracting P γ P γ P_(gamma)P_{\gamma}Pγ from both sides and squaring results in
(3.415) m Λ 2 = P Λ 2 = m Σ 2 − 2 P Σ ⋅ P γ = m Σ ( m Σ − 2 E γ ) (3.415) m Λ 2 = P Λ 2 = m Σ 2 − 2 P Σ ⋅ P γ = m Σ m Σ − 2 E γ {:(3.415)m_(Lambda)^(2)=P_(Lambda)^(2)=m_(Sigma)^(2)-2P_(Sigma)*P_(gamma)=m_(Sigma)(m_(Sigma)-2E_(gamma)):}\begin{equation*} m_{\Lambda}^{2}=P_{\Lambda}^{2}=m_{\Sigma}^{2}-2 P_{\Sigma} \cdot P_{\gamma}=m_{\Sigma}\left(m_{\Sigma}-2 E_{\gamma}\right) \tag{3.415} \end{equation*}(3.415)mΛ2=PΛ2=mΣ2−2PΣ⋅Pγ=mΣ(mΣ−2Eγ)
where E γ E γ E_(gamma)E_{\gamma}Eγ is the energy of the photon in the rest frame of the Σ 0 Σ 0 Sigma^(0)\Sigma^{0}Σ0 particle. Solving for E γ E γ E_(gamma)E_{\gamma}Eγ results in
(3.416) E γ = m Σ 2 − m Λ 2 2 m Σ (3.416) E γ = m Σ 2 − m Λ 2 2 m Σ {:(3.416)E_(gamma)=(m_(Sigma)^(2)-m_(Lambda)^(2))/(2m_(Sigma)):}\begin{equation*} E_{\gamma}=\frac{m_{\Sigma}^{2}-m_{\Lambda}^{2}}{2 m_{\Sigma}} \tag{3.416} \end{equation*}(3.416)Eγ=mΣ2−mΛ22mΣ
c) Since the Σ 0 Σ 0 Sigma^(0)\Sigma^{0}Σ0 particle is moving straight toward the detector, the energy of the photon as registered by the detector will be given by the relativistic Doppler shift formula
(3.417) E γ ′ = E γ 1 + v 1 − v = E γ 1 + 1 / 3 1 − 1 / 3 = E γ 2 = m Σ 2 − m Λ 2 2 m Σ (3.417) E γ ′ = E γ 1 + v 1 − v = E γ 1 + 1 / 3 1 − 1 / 3 = E γ 2 = m Σ 2 − m Λ 2 2 m Σ {:(3.417)E_(gamma)^(')=E_(gamma)sqrt((1+v)/(1-v))=E_(gamma)sqrt((1+1//3)/(1-1//3))=E_(gamma)sqrt2=(m_(Sigma)^(2)-m_(Lambda)^(2))/(sqrt2m_(Sigma)):}\begin{equation*} E_{\gamma}^{\prime}=E_{\gamma} \sqrt{\frac{1+v}{1-v}}=E_{\gamma} \sqrt{\frac{1+1 / 3}{1-1 / 3}}=E_{\gamma} \sqrt{2}=\frac{m_{\Sigma}^{2}-m_{\Lambda}^{2}}{\sqrt{2} m_{\Sigma}} \tag{3.417} \end{equation*}(3.417)Eγ′=Eγ1+v1−v=Eγ1+1/31−1/3=Eγ2=mΣ2−mΛ22mΣ
1.88
a) In the center-of-mass system, we have by definition p e + p p = 0 p e + p p = 0 p_(e)+p_(p)=0\boldsymbol{p}_{e}+\boldsymbol{p}_{p}=\mathbf{0}pe+pp=0 and conservation of 3 -momentum then leads to p e ′ + p p ′ = 0 p e ′ + p p ′ = 0 p_(e)^(')+p_(p)^(')=0\boldsymbol{p}_{e}^{\prime}+\boldsymbol{p}_{p}^{\prime}=\mathbf{0}pe′+pp′=0. Due to conservation of energy, we have for elastic scattering that | p e | = | p p | = | p e ′ | = | p p ′ | ≡ p p e = p p = p e ′ = p p ′ ≡ p |p_(e)|=|p_(p)|=|p_(e)^(')|=|p_(p)^(')|-=p\left|\boldsymbol{p}_{e}\right|=\left|\boldsymbol{p}_{p}\right|=\left|\boldsymbol{p}_{e}^{\prime}\right|=\left|\boldsymbol{p}_{p}^{\prime}\right| \equiv p|pe|=|pp|=|pe′|=|pp′|≡p. Therefore, we have E e = E e ′ E e = E e ′ E_(e)=E_(e)^(')E_{e}=E_{e}^{\prime}Ee=Ee′. Using these results, we find that t = ( p e − p e ′ ) 2 = t = p e − p e ′ 2 = t=(p_(e)-p_(e)^('))^(2)=t=\left(p_{e}-p_{e}^{\prime}\right)^{2}=t=(pe−pe′)2= ( E e − E e ′ ) 2 − ( p e − p e ′ ) 2 = − ( p e − p e ′ ) 2 E e − E e ′ 2 − p e − p e ′ 2 = − p e − p e ′ 2 (E_(e)-E_(e)^('))^(2)-(p_(e)-p_(e)^('))^(2)=-(p_(e)-p_(e)^('))^(2)\left(E_{e}-E_{e}^{\prime}\right)^{2}-\left(\boldsymbol{p}_{e}-\boldsymbol{p}_{e}^{\prime}\right)^{2}=-\left(\boldsymbol{p}_{e}-\boldsymbol{p}_{e}^{\prime}\right)^{2}(Ee−Ee′)2−(pe−pe′)2=−(pe−pe′)2, and thus, we obtain − t = ( p e − p e ′ ) 2 − t = p e − p e ′ 2 -t=(p_(e)-p_(e)^('))^(2)-t=\left(\boldsymbol{p}_{e}-\boldsymbol{p}_{e}^{\prime}\right)^{2}−t=(pe−pe′)2. Introducing the scattering angle θ θ theta\thetaθ by p e ⋅ p e ′ = p 2 cos ⁡ θ p e ⋅ p e ′ = p 2 cos ⁡ θ p_(e)*p_(e)^(')=p^(2)cos theta\boldsymbol{p}_{e} \cdot \boldsymbol{p}_{e}^{\prime}=p^{2} \cos \thetape⋅pe′=p2cos⁡θ, we find that
− t = p e 2 + p e ′ 2 − 2 p e ⋅ p e ′ = p 2 + p 2 − 2 p 2 cos ⁡ θ (3.418) = 2 p 2 ( 1 − cos ⁡ θ ) = 4 p 2 sin 2 ⁡ θ 2 − t = p e 2 + p e ′ 2 − 2 p e ⋅ p e ′ = p 2 + p 2 − 2 p 2 cos ⁡ θ (3.418) = 2 p 2 ( 1 − cos ⁡ θ ) = 4 p 2 sin 2 ⁡ θ 2 {:[-t=p_(e)^(2)+p_(e)^('2)-2p_(e)*p_(e)^(')=p^(2)+p^(2)-2p^(2)cos theta],[(3.418)=2p^(2)(1-cos theta)=4p^(2)sin^(2)((theta)/(2))]:}\begin{align*} -t=\boldsymbol{p}_{e}^{2}+\boldsymbol{p}_{e}^{\prime 2}-2 \boldsymbol{p}_{e} \cdot \boldsymbol{p}_{e}^{\prime} & =p^{2}+p^{2}-2 p^{2} \cos \theta \\ & =2 p^{2}(1-\cos \theta)=4 p^{2} \sin ^{2} \frac{\theta}{2} \tag{3.418} \end{align*}−t=pe2+pe′2−2pe⋅pe′=p2+p2−2p2cos⁡θ(3.418)=2p2(1−cos⁡θ)=4p2sin2⁡θ2
Thus, the result is − t = 4 p 2 sin 2 ⁡ ( θ / 2 ) − t = 4 p 2 sin 2 ⁡ ( θ / 2 ) -t=4p^(2)sin^(2)(theta//2)-t=4 p^{2} \sin ^{2}(\theta / 2)−t=4p2sin2⁡(θ/2).
b) Since conservation of 4-momentum holds, we also have p e − p e ′ = p p ′ − p p p e − p e ′ = p p ′ − p p p_(e)-p_(e)^(')=p_(p)^(')-p_(p)p_{e}-p_{e}^{\prime}=p_{p}^{\prime}-p_{p}pe−pe′=pp′−pp, which means that t = ( p p ′ − p p ) 2 t = p p ′ − p p 2 t=(p_(p)^(')-p_(p))^(2)t=\left(p_{p}^{\prime}-p_{p}\right)^{2}t=(pp′−pp)2. In the laboratory system, we have p p = ( m p , 0 ) p p = m p , 0 p_(p)=(m_(p),0)p_{p}=\left(m_{p}, \mathbf{0}\right)pp=(mp,0) and p p ′ = ( m p + T p ′ , p p ) p p ′ = m p + T p ′ , p p p_(p)^(')=(m_(p)+T_(p)^('),p_(p))p_{p}^{\prime}=\left(m_{p}+T_{p}^{\prime}, \boldsymbol{p}_{p}\right)pp′=(mp+Tp′,pp). Therefore, we find that
t = ( p p ′ − p p ) 2 = p p ′ 2 + p p 2 − 2 p p ′ ⋅ p p = 2 m p 2 − 2 p p ⋅ p p ′ (3.419) = 2 m p 2 − 2 m p ( m p + T p ′ ) = − 2 m p T p ′ t = p p ′ − p p 2 = p p ′ 2 + p p 2 − 2 p p ′ ⋅ p p = 2 m p 2 − 2 p p ⋅ p p ′ (3.419) = 2 m p 2 − 2 m p m p + T p ′ = − 2 m p T p ′ {:[t=(p_(p)^(')-p_(p))^(2)=p_(p)^('2)+p_(p)^(2)-2p_(p)^(')*p_(p)=2m_(p)^(2)-2p_(p)*p_(p)^(')],[(3.419)=2m_(p)^(2)-2m_(p)(m_(p)+T_(p)^('))=-2m_(p)T_(p)^(')]:}\begin{align*} t & =\left(p_{p}^{\prime}-p_{p}\right)^{2}=p_{p}^{\prime 2}+p_{p}^{2}-2 p_{p}^{\prime} \cdot p_{p}=2 m_{p}^{2}-2 p_{p} \cdot p_{p}^{\prime} \\ & =2 m_{p}^{2}-2 m_{p}\left(m_{p}+T_{p}^{\prime}\right)=-2 m_{p} T_{p}^{\prime} \tag{3.419} \end{align*}t=(pp′−pp)2=pp′2+pp2−2pp′⋅pp=2mp2−2pp⋅pp′(3.419)=2mp2−2mp(mp+Tp′)=−2mpTp′
Thus, the result is T p ′ = − t / ( 2 m p ) T p ′ = − t / 2 m p T_(p)^(')=-t//(2m_(p))T_{p}^{\prime}=-t /\left(2 m_{p}\right)Tp′=−t/(2mp).

1.89

The energy of the pion before the collision is E = m π + T E = m π + T E=m_(pi)+TE=m_{\pi}+TE=mπ+T. Now, s 0 = ( m π + m Δ ) 2 s 0 = m π + m Δ 2 s_(0)=(m_(pi)+m_(Delta))^(2)s_{0}=\left(m_{\pi}+m_{\Delta}\right)^{2}s0=(mπ+mΔ)2 gives the minimal center-of-mass energy squared for production of the Δ Δ Delta\DeltaΔ. However, this is also given by s 0 = ( ( E , p ) + ( m p , 0 ) ) 2 = ( E + m p ) 2 − p 2 = ( T + m π + s 0 = ( E , p ) + m p , 0 2 = E + m p 2 − p 2 = T + m π + s_(0)=((E,p)+(m_(p),0))^(2)=(E+m_(p))^(2)-p^(2)=(T+m_(pi)+:}s_{0}=\left((E, \boldsymbol{p})+\left(m_{p}, \mathbf{0}\right)\right)^{2}=\left(E+m_{p}\right)^{2}-\boldsymbol{p}^{2}=\left(T+m_{\pi}+\right.s0=((E,p)+(mp,0))2=(E+mp)2−p2=(T+mπ+ m p ) 2 − p 2 m p 2 − p 2 m_(p))^(2)-p^(2)\left.m_{p}\right)^{2}-\boldsymbol{p}^{2}mp)2−p2. From E 2 = ( m + T ) 2 = m 2 + p 2 E 2 = ( m + T ) 2 = m 2 + p 2 E^(2)=(m+T)^(2)=m^(2)+p^(2)E^{2}=(m+T)^{2}=m^{2}+p^{2}E2=(m+T)2=m2+p2 follows that p 2 = T 2 + 2 m T p 2 = T 2 + 2 m T p^(2)=T^(2)+2mTp^{2}=T^{2}+2 m Tp2=T2+2mT. Thus, the kinetic energy T T TTT of the pion required to create the Δ Δ Delta\DeltaΔ is given by
( m π + m Δ ) 2 = ( T + m π + m p ) 2 − ( T 2 + 2 m π T ) (3.420) ⟹ T = ( m π + m Δ ) 2 − ( m π + m p ) 2 2 m p m π + m Δ 2 = T + m π + m p 2 − T 2 + 2 m π T (3.420) ⟹ T = m π + m Δ 2 − m π + m p 2 2 m p {:[(m_(pi)+m_(Delta))^(2)=(T+m_(pi)+m_(p))^(2)-(T^(2)+2m_(pi)T)],[(3.420)Longrightarrow T=((m_(pi)+m_(Delta))^(2)-(m_(pi)+m_(p))^(2))/(2m_(p))]:}\begin{align*} \left(m_{\pi}+m_{\Delta}\right)^{2} & =\left(T+m_{\pi}+m_{p}\right)^{2}-\left(T^{2}+2 m_{\pi} T\right) \\ \Longrightarrow T & =\frac{\left(m_{\pi}+m_{\Delta}\right)^{2}-\left(m_{\pi}+m_{p}\right)^{2}}{2 m_{p}} \tag{3.420} \end{align*}(mπ+mΔ)2=(T+mπ+mp)2−(T2+2mπT)(3.420)⟹T=(mπ+mΔ)2−(mπ+mp)22mp
1.90
The square of the center-of-mass energy is given by
(3.421) P tot 2 = ( p π + p d ) 2 = ( p p 1 + p p 2 ) 2 (3.421) P tot 2 = p π + p d 2 = p p 1 + p p 2 2 {:(3.421)P_(tot)^(2)=(p_(pi)+p_(d))^(2)=(p_(p_(1))+p_(p_(2)))^(2):}\begin{equation*} P_{\mathrm{tot}}^{2}=\left(p_{\pi}+p_{d}\right)^{2}=\left(p_{p_{1}}+p_{p_{2}}\right)^{2} \tag{3.421} \end{equation*}(3.421)Ptot2=(pπ+pd)2=(pp1+pp2)2
In the case when the pion hits a deuteron at rest, we have
(3.422) p π = ( m π + T π , p π ) and p d = ( m d , 0 ) (3.422) p π = m π + T π , p π  and  p d = m d , 0 {:(3.422)p_(pi)=(m_(pi)+T_(pi),p_(pi))quad" and "quadp_(d)=(m_(d),0):}\begin{equation*} p_{\pi}=\left(m_{\pi}+T_{\pi}, p_{\pi}\right) \quad \text { and } \quad p_{d}=\left(m_{d}, \mathbf{0}\right) \tag{3.422} \end{equation*}(3.422)pπ=(mπ+Tπ,pπ) and pd=(md,0)
This gives the square of the center-of-mass energy, i.e.,
(3.423) P tot 2 = p π 2 + p d 2 + 2 p π ⋅ p d = ( m π + m d ) 2 + 2 m d T π . (3.423) P tot 2 = p π 2 + p d 2 + 2 p π ⋅ p d = m π + m d 2 + 2 m d T π . {:(3.423)P_(tot)^(2)=p_(pi)^(2)+p_(d)^(2)+2p_(pi)*p_(d)=(m_(pi)+m_(d))^(2)+2m_(d)T_(pi).:}\begin{equation*} P_{\mathrm{tot}}^{2}=p_{\pi}^{2}+p_{d}^{2}+2 p_{\pi} \cdot p_{d}=\left(m_{\pi}+m_{d}\right)^{2}+2 m_{d} T_{\pi} . \tag{3.423} \end{equation*}(3.423)Ptot2=pπ2+pd2+2pπ⋅pd=(mπ+md)2+2mdTπ.
Similarly, we have, in the case of one proton hitting another proton at rest,
(3.424) p p 1 = ( m p + T p , p p 1 ) and p p 2 = ( m p , 0 ) , (3.424) p p 1 = m p + T p , p p 1  and  p p 2 = m p , 0 , {:(3.424)p_(p_(1))=(m_(p)+T_(p),p_(p_(1)))quad" and "quadp_(p_(2))=(m_(p),0)",":}\begin{equation*} p_{p_{1}}=\left(m_{p}+T_{p}, \boldsymbol{p}_{p_{1}}\right) \quad \text { and } \quad p_{p_{2}}=\left(m_{p}, \mathbf{0}\right), \tag{3.424} \end{equation*}(3.424)pp1=(mp+Tp,pp1) and pp2=(mp,0),
resulting in
(3.425) P tot 2 = 4 m p 2 + 2 m p T p (3.425) P tot 2 = 4 m p 2 + 2 m p T p {:(3.425)P_(tot)^(2)=4m_(p)^(2)+2m_(p)T_(p):}\begin{equation*} P_{\mathrm{tot}}^{2}=4 m_{p}^{2}+2 m_{p} T_{p} \tag{3.425} \end{equation*}(3.425)Ptot2=4mp2+2mpTp
Requiring the two expressions for the square of the center-of-mass energy to be the same, we obtain
(3.426) T p = ( m π + m d ) 2 2 m p − 2 m p + m d m p T π (3.426) T p = m π + m d 2 2 m p − 2 m p + m d m p T π {:(3.426)T_(p)=((m_(pi)+m_(d))^(2))/(2m_(p))-2m_(p)+(m_(d))/(m_(p))T_(pi):}\begin{equation*} T_{p}=\frac{\left(m_{\pi}+m_{d}\right)^{2}}{2 m_{p}}-2 m_{p}+\frac{m_{d}}{m_{p}} T_{\pi} \tag{3.426} \end{equation*}(3.426)Tp=(mπ+md)22mp−2mp+mdmpTπ
1.91
Consider the reaction π + + n ⟶ K + + Λ π + + n ⟶ K + + Λ pi^(+)+n longrightarrowK^(+)+Lambda\pi^{+}+n \longrightarrow K^{+}+\Lambdaπ++n⟶K++Λ. Let θ = [ π + , K + ] = 90 ∘ θ = π + , K + = 90 ∘ theta=[pi^(+),K^(+)]=90^(@)\theta=\left[\pi^{+}, K^{+}\right]=90^{\circ}θ=[π+,K+]=90∘. Conservation of 4-momentum gives
(3.427) p π + + p n = p K + + p Λ . (3.427) p π + + p n = p K + + p Λ . {:(3.427)p_(pi^(+))+p_(n)=p_(K^(+))+p_(Lambda).:}\begin{equation*} p_{\pi^{+}}+p_{n}=p_{K^{+}}+p_{\Lambda} . \tag{3.427} \end{equation*}(3.427)pπ++pn=pK++pΛ.
The unknown kinematics of the Λ Λ Lambda\LambdaΛ particle can be removed by isolating p Λ p Λ p_(Lambda)p_{\Lambda}pΛ and squaring such that
(3.428) p Λ = p π + + p n − p K + , p Λ 2 = ( p π + + p n − p K + ) 2 (3.429) = p π + 2 + p n 2 + p K + 2 + 2 p π + ⋅ p n − 2 p π + ⋅ p K + − 2 p n ⋅ p K + . (3.428) p Λ = p π + + p n − p K + , p Λ 2 = p π + + p n − p K + 2 (3.429) = p π + 2 + p n 2 + p K + 2 + 2 p π + ⋅ p n − 2 p π + ⋅ p K + − 2 p n ⋅ p K + . {:[(3.428)p_(Lambda)=p_(pi^(+))+p_(n)-p_(K^(+))","],[p_(Lambda)^(2)=(p_(pi^(+))+p_(n)-p_(K^(+)))^(2)],[(3.429)=p_(pi^(+))^(2)+p_(n)^(2)+p_(K^(+))^(2)+2p_(pi^(+))*p_(n)-2p_(pi^(+))*p_(K^(+))-2p_(n)*p_(K^(+)).]:}\begin{align*} p_{\Lambda} & =p_{\pi^{+}}+p_{n}-p_{K^{+}}, \tag{3.428}\\ p_{\Lambda}^{2} & =\left(p_{\pi^{+}}+p_{n}-p_{K^{+}}\right)^{2} \\ & =p_{\pi^{+}}^{2}+p_{n}^{2}+p_{K^{+}}^{2}+2 p_{\pi^{+}} \cdot p_{n}-2 p_{\pi^{+}} \cdot p_{K^{+}}-2 p_{n} \cdot p_{K^{+}} . \tag{3.429} \end{align*}(3.428)pΛ=pπ++pn−pK+,pΛ2=(pπ++pn−pK+)2(3.429)=pπ+2+pn2+pK+2+2pπ+⋅pn−2pπ+⋅pK+−2pn⋅pK+.
Using the fact that p 2 = m 2 p 2 = m 2 p^(2)=m^(2)p^{2}=m^{2}p2=m2 implies that
(3.430) m Λ 2 = m π + 2 + m n 2 + m K + 2 + 2 p π + ⋅ p n − 2 p π + ⋅ p K + − 2 p n ⋅ p K + (3.430) m Λ 2 = m π + 2 + m n 2 + m K + 2 + 2 p π + ⋅ p n − 2 p π + ⋅ p K + − 2 p n ⋅ p K + {:(3.430)m_(Lambda)^(2)=m_(pi^(+))^(2)+m_(n)^(2)+m_(K^(+))^(2)+2p_(pi^(+))*p_(n)-2p_(pi^(+))*p_(K^(+))-2p_(n)*p_(K^(+)):}\begin{equation*} m_{\Lambda}^{2}=m_{\pi^{+}}^{2}+m_{n}^{2}+m_{K^{+}}^{2}+2 p_{\pi^{+}} \cdot p_{n}-2 p_{\pi^{+}} \cdot p_{K^{+}}-2 p_{n} \cdot p_{K^{+}} \tag{3.430} \end{equation*}(3.430)mΛ2=mπ+2+mn2+mK+2+2pπ+⋅pn−2pπ+⋅pK+−2pn⋅pK+
In the rest frame of n n nnn, one has
(3.431) p π + = ( E π + , p π + ) , p n = ( m n , 0 ) , and p K + = ( E K + , p K + ) . (3.431) p π + = E π + , p π + , p n = m n , 0 ,  and  p K + = E K + , p K + . {:(3.431)p_(pi^(+))=(E_(pi^(+)),p_(pi^(+)))","quadp_(n)=(m_(n),0)","quad" and "quadp_(K^(+))=(E_(K^(+)),p_(K^(+))).:}\begin{equation*} p_{\pi^{+}}=\left(E_{\pi^{+}}, \boldsymbol{p}_{\pi^{+}}\right), \quad p_{n}=\left(m_{n}, \mathbf{0}\right), \quad \text { and } \quad p_{K^{+}}=\left(E_{K^{+}}, \boldsymbol{p}_{K^{+}}\right) . \tag{3.431} \end{equation*}(3.431)pπ+=(Eπ+,pπ+),pn=(mn,0), and pK+=(EK+,pK+).
This leads to
(3.432) m Λ 2 = m π + 2 + m n 2 + m K + 2 + 2 E π + ⋅ m n − 2 ( E π + E K + − p π + ⋅ p K + ) − 2 m n ⋅ E K + (3.432) m Λ 2 = m π + 2 + m n 2 + m K + 2 + 2 E π + ⋅ m n − 2 E π + E K + − p π + ⋅ p K + − 2 m n ⋅ E K + {:(3.432)m_(Lambda)^(2)=m_(pi^(+))^(2)+m_(n)^(2)+m_(K^(+))^(2)+2E_(pi^(+))*m_(n)-2(E_(pi^(+))E_(K^(+))-p_(pi^(+))*p_(K^(+)))-2m_(n)*E_(K^(+)):}\begin{equation*} m_{\Lambda}^{2}=m_{\pi^{+}}^{2}+m_{n}^{2}+m_{K^{+}}^{2}+2 E_{\pi^{+}} \cdot m_{n}-2\left(E_{\pi^{+}} E_{K^{+}}-\boldsymbol{p}_{\pi^{+}} \cdot \boldsymbol{p}_{K^{+}}\right)-2 m_{n} \cdot E_{K^{+}} \tag{3.432} \end{equation*}(3.432)mΛ2=mπ+2+mn2+mK+2+2Eπ+⋅mn−2(Eπ+EK+−pπ+⋅pK+)−2mn⋅EK+
Using p π + ⋅ p K + = | p π + | | p K + | cos ⁡ θ = | p π + | | p K + | ⋅ 0 = 0 p π + ⋅ p K + = p π + p K + cos ⁡ θ = p π + p K + ⋅ 0 = 0 p_(pi^(+))*p_(K^(+))=|p_(pi^(+))||p_(K^(+))|cos theta=|p_(pi^(+))||p_(K^(+))|*0=0\boldsymbol{p}_{\pi^{+}} \cdot \boldsymbol{p}_{K^{+}}=\left|\boldsymbol{p}_{\pi^{+}}\right|\left|\boldsymbol{p}_{K^{+}}\right| \cos \theta=\left|\boldsymbol{p}_{\pi^{+}}\right|\left|\boldsymbol{p}_{K^{+}}\right| \cdot 0=0pπ+⋅pK+=|pπ+||pK+|cos⁡θ=|pπ+||pK+|⋅0=0 yields
(3.433) m Λ 2 = m π + 2 + m n 2 + m K + 2 + 2 E π + ⋅ m n − 2 E π + E K + − 2 m n ⋅ E K + (3.433) m Λ 2 = m π + 2 + m n 2 + m K + 2 + 2 E π + ⋅ m n − 2 E π + E K + − 2 m n ⋅ E K + {:(3.433)m_(Lambda)^(2)=m_(pi^(+))^(2)+m_(n)^(2)+m_(K^(+))^(2)+2E_(pi^(+))*m_(n)-2E_(pi^(+))E_(K^(+))-2m_(n)*E_(K^(+)):}\begin{equation*} m_{\Lambda}^{2}=m_{\pi^{+}}^{2}+m_{n}^{2}+m_{K^{+}}^{2}+2 E_{\pi^{+}} \cdot m_{n}-2 E_{\pi^{+}} E_{K^{+}}-2 m_{n} \cdot E_{K^{+}} \tag{3.433} \end{equation*}(3.433)mΛ2=mπ+2+mn2+mK+2+2Eπ+⋅mn−2Eπ+EK+−2mn⋅EK+
Solving the above equation for E π + E π + E_(pi^(+))E_{\pi^{+}}Eπ+, we find that
(3.434) E π + = m Λ 2 − m π + 2 − m n 2 − m K + 2 + 2 E K + m n 2 ( m n − E K + ) (3.434) E π + = m Λ 2 − m π + 2 − m n 2 − m K + 2 + 2 E K + m n 2 m n − E K + {:(3.434)E_(pi^(+))=(m_(Lambda)^(2)-m_(pi^(+))^(2)-m_(n)^(2)-m_(K^(+))^(2)+2E_(K^(+))m_(n))/(2(m_(n)-E_(K^(+)))):}\begin{equation*} E_{\pi^{+}}=\frac{m_{\Lambda}^{2}-m_{\pi^{+}}^{2}-m_{n}^{2}-m_{K^{+}}^{2}+2 E_{K^{+}} m_{n}}{2\left(m_{n}-E_{K^{+}}\right)} \tag{3.434} \end{equation*}(3.434)Eπ+=mΛ2−mπ+2−mn2−mK+2+2EK+mn2(mn−EK+)
The kinetic energy of π + π + pi^(+)\pi^{+}π+is given by T π + = E π + − m π + T π + = E π + − m π + T_(pi^(+))=E_(pi^(+))-m_(pi^(+))T_{\pi^{+}}=E_{\pi^{+}}-m_{\pi^{+}}Tπ+=Eπ+−mπ+. Thus, one obtains
(3.435) T π + = m Λ 2 − m π + 2 − m n 2 − m K + 2 + 2 E K + m n 2 ( m n − E K + ) − m π + (3.435) T π + = m Λ 2 − m π + 2 − m n 2 − m K + 2 + 2 E K + m n 2 m n − E K + − m π + {:(3.435)T_(pi^(+))=(m_(Lambda)^(2)-m_(pi^(+))^(2)-m_(n)^(2)-m_(K^(+))^(2)+2E_(K^(+))m_(n))/(2(m_(n)-E_(K^(+))))-m_(pi^(+)):}\begin{equation*} T_{\pi^{+}}=\frac{m_{\Lambda}^{2}-m_{\pi^{+}}^{2}-m_{n}^{2}-m_{K^{+}}^{2}+2 E_{K^{+}} m_{n}}{2\left(m_{n}-E_{K^{+}}\right)}-m_{\pi^{+}} \tag{3.435} \end{equation*}(3.435)Tπ+=mΛ2−mπ+2−mn2−mK+2+2EK+mn2(mn−EK+)−mπ+
Using the given quantities T T TTT and E E EEE, i.e., T π + = T T π + = T T_(pi^(+))=TT_{\pi^{+}}=TTπ+=T and E K + = E E K + = E E_(K^(+))=EE_{K^{+}}=EEK+=E, this gives the result
(3.436) T = m Λ 2 − m π + 2 − m n 2 − m K + 2 + 2 E m n 2 ( m n − E ) − m π + (3.436) T = m Λ 2 − m π + 2 − m n 2 − m K + 2 + 2 E m n 2 m n − E − m π + {:(3.436)T=(m_(Lambda)^(2)-m_(pi^(+))^(2)-m_(n)^(2)-m_(K^(+))^(2)+2Em_(n))/(2(m_(n)-E))-m_(pi^(+)):}\begin{equation*} T=\frac{m_{\Lambda}^{2}-m_{\pi^{+}}^{2}-m_{n}^{2}-m_{K^{+}}^{2}+2 E m_{n}}{2\left(m_{n}-E\right)}-m_{\pi^{+}} \tag{3.436} \end{equation*}(3.436)T=mΛ2−mπ+2−mn2−mK+2+2Emn2(mn−E)−mπ+

1.92

Let the 4-momenta of the different particles be indexed by their respective symbols. Then, conservation of 4-momentum gives
(3.437) p p + p π − = p n + p π 0 (3.437) p p + p π − = p n + p π 0 {:(3.437)p_(p)+p_(pi^(-))=p_(n)+p_(pi^(0)):}\begin{equation*} p_{p}+p_{\pi^{-}}=p_{n}+p_{\pi^{0}} \tag{3.437} \end{equation*}(3.437)pp+pπ−=pn+pπ0
Since we have no information on the π 0 π 0 pi^(0)\pi^{0}π0 meson, we solve for its 4 -momentum and square it, which leads to
(3.438) m π 0 2 = ( p p + p π − − p n ) 2 (3.438) m π 0 2 = p p + p π − − p n 2 {:(3.438)m_(pi^(0))^(2)=(p_(p)+p_(pi^(-))-p_(n))^(2):}\begin{equation*} m_{\pi^{0}}^{2}=\left(p_{p}+p_{\pi^{-}}-p_{n}\right)^{2} \tag{3.438} \end{equation*}(3.438)mπ02=(pp+pπ−−pn)2
Thus, in the common rest frame of the incoming particles, we obtain
(3.439) m π 0 = ( m p + m π − − E n ) 2 − p n 2 = ( m p + m π − ) 2 + m n 2 − 2 E n ( m p + m π − ) (3.439) m π 0 = m p + m π − − E n 2 − p n 2 = m p + m π − 2 + m n 2 − 2 E n m p + m π − {:(3.439)m_(pi^(0))=sqrt((m_(p)+m_(pi^(-))-E_(n))^(2)-p_(n)^(2))=sqrt((m_(p)+m_(pi^(-)))^(2)+m_(n)^(2)-2E_(n)(m_(p)+m_(pi^(-)))):}\begin{equation*} m_{\pi^{0}}=\sqrt{\left(m_{p}+m_{\pi^{-}}-E_{n}\right)^{2}-p_{n}^{2}}=\sqrt{\left(m_{p}+m_{\pi^{-}}\right)^{2}+m_{n}^{2}-2 E_{n}\left(m_{p}+m_{\pi^{-}}\right)} \tag{3.439} \end{equation*}(3.439)mπ0=(mp+mπ−−En)2−pn2=(mp+mπ−)2+mn2−2En(mp+mπ−)
Now, we have v n ≈ 3 ⋅ 10 − 2 v n ≈ 3 ⋅ 10 − 2 v_(n)~~3*10^(-2)v_{n} \approx 3 \cdot 10^{-2}vn≈3⋅10−2, which is small compared to 1 . Therefore, we can approximate
(3.440) E n = m n 1 − v n 2 ≃ m n ( 1 + 1 2 ⋅ v n 2 ) (3.440) E n = m n 1 − v n 2 ≃ m n 1 + 1 2 ⋅ v n 2 {:(3.440)E_(n)=(m_(n))/(sqrt(1-v_(n)^(2)))≃m_(n)(1+(1)/(2)*v_(n)^(2)):}\begin{equation*} E_{n}=\frac{m_{n}}{\sqrt{1-v_{n}^{2}}} \simeq m_{n}\left(1+\frac{1}{2} \cdot v_{n}^{2}\right) \tag{3.440} \end{equation*}(3.440)En=mn1−vn2≃mn(1+12⋅vn2)
which yields
m π 0 ≃ ( m p + m π − ) 2 + m n 2 − 2 ( m p + m π − ) m n ( 1 + 1 2 v n 2 ) = ( m p + m π − − m n ) 2 − m n ( m p + m π − ) v n 2 (3.441) = ( m p + m π − − m n ) 2 1 − m n ( m p + m π − ) ( m p + m π − − m n ) 2 v n 2 m π 0 ≃ m p + m π − 2 + m n 2 − 2 m p + m π − m n 1 + 1 2 v n 2 = m p + m π − − m n 2 − m n m p + m π − v n 2 (3.441) = m p + m π − − m n 2 1 − m n m p + m π − m p + m π − − m n 2 v n 2 {:[m_(pi^(0))≃sqrt((m_(p)+m_(pi^(-)))^(2)+m_(n)^(2)-2(m_(p)+m_(pi^(-)))m_(n)(1+(1)/(2)v_(n)^(2)))],[=sqrt((m_(p)+m_(pi^(-))-m_(n))^(2)-m_(n)(m_(p)+m_(pi^(-)))v_(n)^(2))],[(3.441)=(m_(p)+m_(pi^(-))-m_(n))^(2)sqrt(1-(m_(n)(m_(p)+m_(pi^(-))))/((m_(p)+m_(pi^(-))-m_(n))^(2)))v_(n)^(2)]:}\begin{align*} m_{\pi^{0}} & \simeq \sqrt{\left(m_{p}+m_{\pi^{-}}\right)^{2}+m_{n}^{2}-2\left(m_{p}+m_{\pi^{-}}\right) m_{n}\left(1+\frac{1}{2} v_{n}^{2}\right)} \\ & =\sqrt{\left(m_{p}+m_{\pi^{-}}-m_{n}\right)^{2}-m_{n}\left(m_{p}+m_{\pi^{-}}\right) v_{n}^{2}} \\ & =\left(m_{p}+m_{\pi^{-}}-m_{n}\right)^{2} \sqrt{1-\frac{m_{n}\left(m_{p}+m_{\pi^{-}}\right)}{\left(m_{p}+m_{\pi^{-}}-m_{n}\right)^{2}}} v_{n}^{2} \tag{3.441} \end{align*}mπ0≃(mp+mπ−)2+mn2−2(mp+mπ−)mn(1+12vn2)=(mp+mπ−−mn)2−mn(mp+mπ−)vn2(3.441)=(mp+mπ−−mn)21−mn(mp+mπ−)(mp+mπ−−mn)2vn2
and finally, we find that
(3.442) m π 0 ≃ m p + m π − − m n − 1 2 m n ( m p + m π − ) m p + m π − − m n v n 2 (3.442) m π 0 ≃ m p + m π − − m n − 1 2 m n m p + m π − m p + m π − − m n v n 2 {:(3.442)m_(pi^(0))≃m_(p)+m_(pi^(-))-m_(n)-(1)/(2)(m_(n)(m_(p)+m_(pi^(-))))/(m_(p)+m_(pi^(-))-m_(n))v_(n)^(2):}\begin{equation*} m_{\pi^{0}} \simeq m_{p}+m_{\pi^{-}}-m_{n}-\frac{1}{2} \frac{m_{n}\left(m_{p}+m_{\pi^{-}}\right)}{m_{p}+m_{\pi^{-}}-m_{n}} v_{n}^{2} \tag{3.442} \end{equation*}(3.442)mπ0≃mp+mπ−−mn−12mn(mp+mπ−)mp+mπ−−mnvn2

1.93

The kinetic energy of the neutron is negligible (the kinetic energy of a thermal neutron is of the order 25 meV ), so both the neutron and the proton can be considered at rest before the reaction. By conservation of 4-momentum, we have the relation
(3.443) p p + p n − p γ = p d ⇒ ( p p + p n ) 2 − 2 ( p p + p n ) ⋅ p γ = p d 2 (3.443) p p + p n − p γ = p d ⇒ p p + p n 2 − 2 p p + p n ⋅ p γ = p d 2 {:(3.443)p_(p)+p_(n)-p_(gamma)=p_(d)quad=>quad(p_(p)+p_(n))^(2)-2(p_(p)+p_(n))*p_(gamma)=p_(d)^(2):}\begin{equation*} p_{p}+p_{n}-p_{\gamma}=p_{d} \quad \Rightarrow \quad\left(p_{p}+p_{n}\right)^{2}-2\left(p_{p}+p_{n}\right) \cdot p_{\gamma}=p_{d}^{2} \tag{3.443} \end{equation*}(3.443)pp+pn−pγ=pd⇒(pp+pn)2−2(pp+pn)⋅pγ=pd2
If we let M = m p + m n M = m p + m n M=m_(p)+m_(n)M=m_{p}+m_{n}M=mp+mn, then p p + p n = ( M , 0 ) p p + p n = ( M , 0 ) p_(p)+p_(n)=(M,0)p_{p}+p_{n}=(M, \mathbf{0})pp+pn=(M,0) and p γ = ( E γ , p γ ) p γ = E γ , p γ p_(gamma)=(E_(gamma),p_(gamma))p_{\gamma}=\left(E_{\gamma}, \boldsymbol{p}_{\gamma}\right)pγ=(Eγ,pγ) in the lab frame, while p d = ( M − B , 0 ) p d = ( M − B , 0 ) p_(d)=(M-B,0)p_{d}=(M-B, 0)pd=(M−B,0) in the rest frame of the deutron. It follows from the Lorentz invariance of the Minkowski product that
M 2 − 2 M E γ = M 2 − 2 M B + B 2 (3.444) ⇒ E γ = B ( 1 − B 2 M ) = B ( 1 − B 2 ( m p + m n ) ) M 2 − 2 M E γ = M 2 − 2 M B + B 2 (3.444) ⇒ E γ = B 1 − B 2 M = B 1 − B 2 m p + m n {:[M^(2)-2ME_(gamma)=M^(2)-2MB+B^(2)],[(3.444)quad=>quadE_(gamma)=B(1-(B)/(2M))=B(1-(B)/(2(m_(p)+m_(n))))]:}\begin{align*} & M^{2}-2 M E_{\gamma}=M^{2}-2 M B+B^{2} \\ & \quad \Rightarrow \quad E_{\gamma}=B\left(1-\frac{B}{2 M}\right)=B\left(1-\frac{B}{2\left(m_{p}+m_{n}\right)}\right) \tag{3.444} \end{align*}M2−2MEγ=M2−2MB+B2(3.444)⇒Eγ=B(1−B2M)=B(1−B2(mp+mn))
The difference between the energies B B BBB and E γ E γ E_(gamma)E_{\gamma}Eγ is due to the recoil energy of the deutron.

1.94

Let the 4 -momenta for the particles in the reaction be k , p H , p p k , p H , p p k,p_(H),p_(p)k, p_{\mathrm{H}}, p_{p}k,pH,pp, and p e p e p_(e)p_{e}pe. Since 4 -momentum is conserved, we have k + p H = p p + p e k + p H = p p + p e k+p_(H)=p_(p)+p_(e)k+p_{\mathrm{H}}=p_{p}+p_{e}k+pH=pp+pe. From Lorentz invariance, we also have ( k + p H ) 2 = ( p p + p e ) 2 k + p H 2 = p p + p e 2 (k+p_(H))^(2)=(p_(p)+p_(e))^(2)\left(k+p_{\mathrm{H}}\right)^{2}=\left(p_{p}+p_{e}\right)^{2}(k+pH)2=(pp+pe)2. Now, since both sides of this relation are Lorentz invariants, we can calculate them in different inertial systems. In the system where H is at rest, we can take k = ( ω , ω , 0 , 0 ) k = ( ω , ω , 0 , 0 ) k=(omega,omega,0,0)k=(\omega, \omega, 0,0)k=(ω,ω,0,0) and p H = ( m H , 0 , 0 , 0 ) p H = m H , 0 , 0 , 0 p_(H)=(m_(H),0,0,0)p_{\mathrm{H}}=\left(m_{\mathrm{H}}, 0,0,0\right)pH=(mH,0,0,0). Thus, we obtain ( k + p H ) 2 = 2 ω m H + m H 2 k + p H 2 = 2 ω m H + m H 2 (k+p_(H))^(2)=2omegam_(H)+m_(H)^(2)\left(k+p_{\mathrm{H}}\right)^{2}=2 \omega m_{\mathrm{H}}+m_{\mathrm{H}}^{2}(k+pH)2=2ωmH+mH2. For the final particles, we choose the center-ofmass frame, where the particles are at rest at threshold for the reaction to occur.
Thus, we find that ( p p + p e ) 2 = ( m p + m e ) 2 = ( m H + B ) 2 p p + p e 2 = m p + m e 2 = m H + B 2 (p_(p)+p_(e))^(2)=(m_(p)+m_(e))^(2)=(m_(H)+B)^(2)\left(p_{p}+p_{e}\right)^{2}=\left(m_{p}+m_{e}\right)^{2}=\left(m_{\mathrm{H}}+B\right)^{2}(pp+pe)2=(mp+me)2=(mH+B)2. Finally, combining the above expressions and solving for ω ω omega\omegaω, we obtain
(3.445) ω = B ( 1 + B 2 m H ) (3.445) ω = B 1 + B 2 m H {:(3.445)omega=B(1+(B)/(2m_(H))):}\begin{equation*} \omega=B\left(1+\frac{B}{2 m_{\mathrm{H}}}\right) \tag{3.445} \end{equation*}(3.445)ω=B(1+B2mH)
1.95
In order for the collision of the two neutrinos to produce a Z 0 Z 0 Z^(0)Z^{0}Z0 boson, the center-ofmomentum energy must be equal to the Z 0 Z 0 Z^(0)Z^{0}Z0 mass, i.e.,
(3.446) ( p 1 + p 2 ) 2 = m Z 0 2 (3.446) p 1 + p 2 2 = m Z 0 2 {:(3.446)(p_(1)+p_(2))^(2)=m_(Z^(0))^(2):}\begin{equation*} \left(p_{1}+p_{2}\right)^{2}=m_{Z^{0}}^{2} \tag{3.446} \end{equation*}(3.446)(p1+p2)2=mZ02
where p 1 p 1 p_(1)p_{1}p1 is the 4 -momentum of the CNB neutrino and p 2 p 2 p_(2)p_{2}p2 is the 4 -momentum of the UHE neutrino. The two 4-momenta are given by
(3.447) p 1 = ( m v + E k , E k ( E k + 2 m μ ) a ) (3.448) p 2 = ( E , E b ) (3.447) p 1 = m v + E k , E k E k + 2 m μ a (3.448) p 2 = ( E , E b ) {:[(3.447)p_(1)=(m_(v)+E_(k),sqrt(E_(k)(E_(k)+2m_(mu)))a)],[(3.448)p_(2)=(E","Eb)]:}\begin{align*} & p_{1}=\left(m_{v}+E_{k}, \sqrt{E_{k}\left(E_{k}+2 m_{\mu}\right)} \boldsymbol{a}\right) \tag{3.447}\\ & p_{2}=(E, E \boldsymbol{b}) \tag{3.448} \end{align*}(3.447)p1=(mv+Ek,Ek(Ek+2mμ)a)(3.448)p2=(E,Eb)
where m v m v m_(v)m_{v}mv is the neutrino mass, E k E k E_(k)E_{k}Ek is the thermal energy of the CNB neutrino, E E EEE is the energy of the UHE neutrino, and a a a\boldsymbol{a}a and b b b\boldsymbol{b}b are 3-vectors with modulus one ( | a | = | b | = 1 ) ( | a | = | b | = 1 ) (|a|=|b|=1)(|\boldsymbol{a}|=|\boldsymbol{b}|=1)(|a|=|b|=1). Note that we have neglected the neutrino mass compared to the UHE neutrino energy. In general, we now need to take care if we obtain an expression, where terms including E E EEE cancel (since the corrections of the small parameter m μ / E m μ / E m_(mu)//Em_{\mu} / Emμ/E would then become the leading terms). However, this will not be the case in this problem.
Summing and then squaring the 4 -vectors results in
m Z 0 2 = ( p 1 + p 2 ) 2 = 2 p 1 ⋅ p 2 + p 1 2 + p 2 2 ⏟ ≪ p 1 ⋅ p 2 ≃ 2 p 1 ⋅ p 2 (3.449) = E [ m v + E k − E k ( E k + 2 m v ) a ⋅ b ] m Z 0 2 = p 1 + p 2 2 = 2 p 1 ⋅ p 2 + p 1 2 + p 2 2 ⏟ ≪ p 1 ⋅ p 2 ≃ 2 p 1 ⋅ p 2 (3.449) = E m v + E k − E k E k + 2 m v a ⋅ b {:[m_(Z^(0))^(2)=(p_(1)+p_(2))^(2)=2p_(1)*p_(2)+ubrace(p_(1)^(2)+p_(2)^(2)ubrace)_(≪p_(1)*p_(2))≃2p_(1)*p_(2)],[(3.449)=E[m_(v)+E_(k)-sqrt(E_(k)(E_(k)+2m_(v)))a*b]]:}\begin{align*} m_{Z^{0}}^{2}=\left(p_{1}+p_{2}\right)^{2} & =2 p_{1} \cdot p_{2}+\underbrace{p_{1}^{2}+p_{2}^{2}}_{\ll p_{1} \cdot p_{2}} \simeq 2 p_{1} \cdot p_{2} \\ & =E\left[m_{v}+E_{k}-\sqrt{E_{k}\left(E_{k}+2 m_{v}\right)} \boldsymbol{a} \cdot \boldsymbol{b}\right] \tag{3.449} \end{align*}mZ02=(p1+p2)2=2p1⋅p2+p12+p22⏟≪p1⋅p2≃2p1⋅p2(3.449)=E[mv+Ek−Ek(Ek+2mv)a⋅b]
Furthermore, since | a ⋅ b | ≤ 1 | a ⋅ b | ≤ 1 |a*b| <= 1|\boldsymbol{a} \cdot \boldsymbol{b}| \leq 1|a⋅b|≤1, we obtain
m Z 0 2 ≤ E [ m v + E k − E k ( E k + 2 m v ) ] (3.450) ⇒ E ≥ m Z 0 2 2 1 m v + E k + E k ( E k + 2 m v ) m Z 0 2 ≤ E m v + E k − E k E k + 2 m v (3.450) ⇒ E ≥ m Z 0 2 2 1 m v + E k + E k E k + 2 m v {:[m_(Z^(0))^(2) <= E[m_(v)+E_(k)-sqrt(E_(k)(E_(k)+2m_(v)))]],[(3.450)quad=>quad E >= (m_(Z^(0))^(2))/(2)(1)/(m_(v)+E_(k)+sqrt(E_(k)(E_(k)+2m_(v))))]:}\begin{align*} & m_{Z^{0}}^{2} \leq E\left[m_{v}+E_{k}-\sqrt{E_{k}\left(E_{k}+2 m_{v}\right)}\right] \\ & \quad \Rightarrow \quad E \geq \frac{m_{Z^{0}}^{2}}{2} \frac{1}{m_{v}+E_{k}+\sqrt{E_{k}\left(E_{k}+2 m_{v}\right)}} \tag{3.450} \end{align*}mZ02≤E[mv+Ek−Ek(Ek+2mv)](3.450)⇒E≥mZ0221mv+Ek+Ek(Ek+2mv)
For the case when m v = 0.15 eV ≫ E k m v = 0.15 eV ≫ E k m_(v)=0.15eV≫E_(k)m_{v}=0.15 \mathrm{eV} \gg E_{k}mv=0.15eV≫Ek, this simplifies to
(3.451) E ≥ m Z 0 2 2 m v ≃ 28 ⋅ 10 21 eV = 28 ZeV (3.451) E ≥ m Z 0 2 2 m v ≃ 28 ⋅ 10 21 eV = 28 ZeV {:(3.451)E >= (m_(Z^(0))^(2))/(2m_(v))≃28*10^(21)eV=28ZeV:}\begin{equation*} E \geq \frac{m_{Z^{0}}^{2}}{2 m_{v}} \simeq 28 \cdot 10^{21} \mathrm{eV}=28 \mathrm{ZeV} \tag{3.451} \end{equation*}(3.451)E≥mZ022mv≃28⋅1021eV=28ZeV
Similarly, in the case when E k ≃ 3 k B T / 2 ≫ m ν E k ≃ 3 k B T / 2 ≫ m ν E_(k)≃3k_(B)T//2≫m_(nu)E_{k} \simeq 3 k_{B} T / 2 \gg m_{\nu}Ek≃3kBT/2≫mν, we have
(3.452) E ≥ m Z 0 2 4 E k ≃ 8 ⋅ 10 24 eV = 8 YeV (3.452) E ≥ m Z 0 2 4 E k ≃ 8 ⋅ 10 24 eV = 8 YeV {:(3.452)E >= (m_(Z^(0))^(2))/(4E_(k))≃8*10^(24)eV=8YeV:}\begin{equation*} E \geq \frac{m_{Z^{0}}^{2}}{4 E_{k}} \simeq 8 \cdot 10^{24} \mathrm{eV}=8 \mathrm{YeV} \tag{3.452} \end{equation*}(3.452)E≥mZ024Ek≃8⋅1024eV=8YeV

1.96

a) Conservation of 4 -momentum yields k + p = k ′ + p ′ k + p = k ′ + p ′ k+p=k^(')+p^(')k+p=k^{\prime}+p^{\prime}k+p=k′+p′. We can remove the redundant information about p ′ p ′ p^(')p^{\prime}p′ as follows
(3.453) ( p + k − k ′ ) 2 = p ′ 2 = m 2 (3.453) p + k − k ′ 2 = p ′ 2 = m 2 {:(3.453)(p+k-k^('))^(2)=p^('2)=m^(2):}\begin{equation*} \left(p+k-k^{\prime}\right)^{2}=p^{\prime 2}=m^{2} \tag{3.453} \end{equation*}(3.453)(p+k−k′)2=p′2=m2
Using k 2 = k ′ 2 = 0 k 2 = k ′ 2 = 0 k^(2)=k^('2)=0k^{2}=k^{\prime 2}=0k2=k′2=0 and simplifying gives
(3.454) p ⋅ k − p ⋅ k ′ − k ⋅ k ′ = 0 (3.454) p ⋅ k − p ⋅ k ′ − k ⋅ k ′ = 0 {:(3.454)p*k-p*k^(')-k*k^(')=0:}\begin{equation*} p \cdot k-p \cdot k^{\prime}-k \cdot k^{\prime}=0 \tag{3.454} \end{equation*}(3.454)p⋅k−p⋅k′−k⋅k′=0
Inserting p = ( m , 0 ) , k = ( ω , k ) p = ( m , 0 ) , k = ( ω , k ) p=(m,0),k=(omega,k)p=(m, \mathbf{0}), k=(\omega, \boldsymbol{k})p=(m,0),k=(ω,k), and k ′ = ( ω ′ , k ′ ) k ′ = ω ′ , k ′ k^(')=(omega^('),k^('))k^{\prime}=\left(\omega^{\prime}, \boldsymbol{k}^{\prime}\right)k′=(ω′,k′), this leads to
(3.455) ω − ω ′ = 2 m ⋅ ω ω ′ sin 2 ⁡ θ 2 , (3.455) ω − ω ′ = 2 m ⋅ ω ω ′ sin 2 ⁡ θ 2 , {:(3.455)omega-omega^(')=(2)/(m)*omegaomega^(')sin^(2)((theta)/(2))",":}\begin{equation*} \omega-\omega^{\prime}=\frac{2}{m} \cdot \omega \omega^{\prime} \sin ^{2} \frac{\theta}{2}, \tag{3.455} \end{equation*}(3.455)ω−ω′=2m⋅ωω′sin2⁡θ2,
in the rest frame of the initial electron. We then use ω = 2 π ν = { ν = 1 / λ } = 2 π / λ ω = 2 π ν = { ν = 1 / λ } = 2 π / λ omega=2pi nu={nu=1//lambda}=2pi//lambda\omega=2 \pi \nu=\{\nu=1 / \lambda\}=2 \pi / \lambdaω=2πν={ν=1/λ}=2π/λ and similar for ω ′ ω ′ omega^(')\omega^{\prime}ω′ to obtain the Compton formula, i.e.,
(3.456) λ ′ − λ = 4 π m sin 2 ⁡ θ 2 (3.456) λ ′ − λ = 4 π m sin 2 ⁡ θ 2 {:(3.456)lambda^(')-lambda=(4pi)/(m)sin^(2)((theta)/(2)):}\begin{equation*} \lambda^{\prime}-\lambda=\frac{4 \pi}{m} \sin ^{2} \frac{\theta}{2} \tag{3.456} \end{equation*}(3.456)λ′−λ=4πmsin2⁡θ2
b) We use the Mandelstam variable s s sss defined as
(3.457) s ≡ ( p + k ) 2 = p 2 + k 2 + 2 p ⋅ k = m 2 + 2 m ω (3.457) s ≡ ( p + k ) 2 = p 2 + k 2 + 2 p ⋅ k = m 2 + 2 m ω {:(3.457)s-=(p+k)^(2)=p^(2)+k^(2)+2p*k=m^(2)+2m omega:}\begin{equation*} s \equiv(p+k)^{2}=p^{2}+k^{2}+2 p \cdot k=m^{2}+2 m \omega \tag{3.457} \end{equation*}(3.457)s≡(p+k)2=p2+k2+2p⋅k=m2+2mω
On the other hand, due to conservation of 4-momentum, we also have, in the center-of-mass system (i.e., p ′ + k ′ = 0 p ′ + k ′ = 0 p^(')+k^(')=0\boldsymbol{p}^{\prime}+\boldsymbol{k}^{\prime}=\mathbf{0}p′+k′=0 ), the expression
(3.458) s = ( p ′ + k ′ ) 2 = ( E ′ + ω ′ ) 2 = ( m 2 + ω ′ 2 + ω ′ ) 2 (3.458) s = p ′ + k ′ 2 = E ′ + ω ′ 2 = m 2 + ω ′ 2 + ω ′ 2 {:(3.458)s=(p^(')+k^('))^(2)=(E^(')+omega^('))^(2)=(sqrt(m^(2)+omega^('2))+omega^('))^(2):}\begin{equation*} s=\left(p^{\prime}+k^{\prime}\right)^{2}=\left(E^{\prime}+\omega^{\prime}\right)^{2}=\left(\sqrt{m^{2}+\omega^{\prime 2}}+\omega^{\prime}\right)^{2} \tag{3.458} \end{equation*}(3.458)s=(p′+k′)2=(E′+ω′)2=(m2+ω′2+ω′)2
where E ′ = m 2 + p ′ 2 E ′ = m 2 + p ′ 2 E^(')=sqrt(m^(2)+p^('2))E^{\prime}=\sqrt{m^{2}+\boldsymbol{p}^{\prime 2}}E′=m2+p′2 is the energy of the outgoing electron and p ′ 2 = k ′ 2 = ω ′ 2 p ′ 2 = k ′ 2 = ω ′ 2 p^('2)=k^('2)=omega^('2)\boldsymbol{p}^{\prime 2}=\boldsymbol{k}^{\prime 2}=\omega^{\prime 2}p′2=k′2=ω′2. Now, since s s sss is Lorentz invariant, equating the two expressions for s s sss and solving for ω ′ ω ′ omega^(')\omega^{\prime}ω′, we obtain
(3.459) m 2 + 2 m ω = ( m 2 + ω ′ 2 + ω ′ ) 2 ⇔ ω ′ = ω 1 + 2 ω m (3.459) m 2 + 2 m ω = m 2 + ω ′ 2 + ω ′ 2 ⇔ ω ′ = ω 1 + 2 ω m {:(3.459)m^(2)+2m omega=(sqrt(m^(2)+omega^('2))+omega^('))^(2)quad<=>quadomega^(')=(omega)/(sqrt(1+(2omega)/(m))):}\begin{equation*} m^{2}+2 m \omega=\left(\sqrt{m^{2}+\omega^{\prime 2}}+\omega^{\prime}\right)^{2} \quad \Leftrightarrow \quad \omega^{\prime}=\frac{\omega}{\sqrt{1+\frac{2 \omega}{m}}} \tag{3.459} \end{equation*}(3.459)m2+2mω=(m2+ω′2+ω′)2⇔ω′=ω1+2ωm

1.97

Denoting the incoming 4-momenta of the electron and photon p e p e p_(e)p_{e}pe and p γ p γ p_(gamma)p_{\gamma}pγ, respectively, and the corresponding outgoing quantities by k e k e k_(e)k_{e}ke and k γ k γ k_(gamma)k_{\gamma}kγ, we have
(3.460) p e = ( m e , 0 ) , p γ = ω ( 1 , 1 , 0 ) , k e = ( E e , p e ) , k γ = ω ′ ( 1 , cos ⁡ θ , sin ⁡ θ ) (3.460) p e = m e , 0 , p γ = ω ( 1 , 1 , 0 ) , k e = E e , p e , k γ = ω ′ ( 1 , cos ⁡ θ , sin ⁡ θ ) {:(3.460)p_(e)=(m_(e),0)","p_(gamma)=omega(1","1","0)","k_(e)=(E_(e),p_(e))","k_(gamma)=omega^(')(1","cos theta","sin theta):}\begin{equation*} p_{e}=\left(m_{e}, \mathbf{0}\right), p_{\gamma}=\omega(1,1,0), k_{e}=\left(E_{e}, \boldsymbol{p}_{e}\right), k_{\gamma}=\omega^{\prime}(1, \cos \theta, \sin \theta) \tag{3.460} \end{equation*}(3.460)pe=(me,0),pγ=ω(1,1,0),ke=(Ee,pe),kγ=ω′(1,cos⁡θ,sin⁡θ)
in the laboratory frame, where we have oriented our coordinate system such that there is no momentum in the z z zzz-direction either before or after the collision. Thus, we have also omitted the z z zzz-components of the 4 -vectors. Conservation of 4-momentum yields
(3.461) p e + p γ = k e + k γ ⟹ p e + p γ − k γ = p e . (3.461) p e + p γ = k e + k γ ⟹ p e + p γ − k γ = p e . {:(3.461)p_(e)+p_(gamma)=k_(e)+k_(gamma)quad Longrightarrowquadp_(e)+p_(gamma)-k_(gamma)=p_(e).:}\begin{equation*} p_{e}+p_{\gamma}=k_{e}+k_{\gamma} \quad \Longrightarrow \quad p_{e}+p_{\gamma}-k_{\gamma}=p_{e} . \tag{3.461} \end{equation*}(3.461)pe+pγ=ke+kγ⟹pe+pγ−kγ=pe.
Squaring this relation, we obtain
m 2 + 2 p e ⋅ p γ − 2 p e ⋅ k γ − 2 p γ ⋅ k γ = m 2 ⟹ m ω = m ω ′ + ω ω ′ ( 1 − cos ⁡ θ ) m 2 + 2 p e ⋅ p γ − 2 p e ⋅ k γ − 2 p γ ⋅ k γ = m 2 ⟹ m ω = m ω ′ + ω ω ′ ( 1 − cos ⁡ θ ) m^(2)+2p_(e)*p_(gamma)-2p_(e)*k_(gamma)-2p_(gamma)*k_(gamma)=m^(2)Longrightarrow m omega=momega^(')+omegaomega^(')(1-cos theta)m^{2}+2 p_{e} \cdot p_{\gamma}-2 p_{e} \cdot k_{\gamma}-2 p_{\gamma} \cdot k_{\gamma}=m^{2} \Longrightarrow m \omega=m \omega^{\prime}+\omega \omega^{\prime}(1-\cos \theta)m2+2pe⋅pγ−2pe⋅kγ−2pγ⋅kγ=m2⟹mω=mω′+ωω′(1−cos⁡θ).
Solving for ω ′ ω ′ omega^(')\omega^{\prime}ω′, we have
(3.463) ω ′ = ω 1 + ω m ( 1 − cos ⁡ θ ) (3.463) ω ′ = ω 1 + ω m ( 1 − cos ⁡ θ ) {:(3.463)omega^(')=(omega)/(1+(omega )/(m)(1-cos theta)):}\begin{equation*} \omega^{\prime}=\frac{\omega}{1+\frac{\omega}{m}(1-\cos \theta)} \tag{3.463} \end{equation*}(3.463)ω′=ω1+ωm(1−cos⁡θ)
Furthermore, the total energy of the outgoing electron E e E e E_(e)E_{e}Ee is given by E e = m + E e = m + E_(e)=m+E_{e}=m+Ee=m+ ω − ω ′ ω − ω ′ omega-omega^(')\omega-\omega^{\prime}ω−ω′ from the time component of the 4-momentum conservation (i.e., energy conservation). Since the kinetic energy is T e = E e − m = ω − ω ′ T e = E e − m = ω − ω ′ T_(e)=E_(e)-m=omega-omega^(')T_{e}=E_{e}-m=\omega-\omega^{\prime}Te=Ee−m=ω−ω′, we insert our result for ω ′ ω ′ omega^(')\omega^{\prime}ω′ and simplify
(3.464) T e = ω − ω ′ = ω [ 1 − 1 1 + ω m ( 1 − cos ⁡ θ ) ] = ω 2 ( 1 − cos ⁡ θ ) m + ω ( 1 − cos ⁡ θ ) (3.464) T e = ω − ω ′ = ω 1 − 1 1 + ω m ( 1 − cos ⁡ θ ) = ω 2 ( 1 − cos ⁡ θ ) m + ω ( 1 − cos ⁡ θ ) {:(3.464)T_(e)=omega-omega^(')=omega[1-(1)/(1+(omega )/(m)(1-cos theta))]=(omega^(2)(1-cos theta))/(m+omega(1-cos theta)):}\begin{equation*} T_{e}=\omega-\omega^{\prime}=\omega\left[1-\frac{1}{1+\frac{\omega}{m}(1-\cos \theta)}\right]=\frac{\omega^{2}(1-\cos \theta)}{m+\omega(1-\cos \theta)} \tag{3.464} \end{equation*}(3.464)Te=ω−ω′=ω[1−11+ωm(1−cos⁡θ)]=ω2(1−cos⁡θ)m+ω(1−cos⁡θ)

1.98

The setup is shown in Figure 3.8. The 4-momenta of the particles are given by (suppressing the z z zzz-component):
(3.465) k = E 0 ( 1 , 1 , 0 ) , p = m γ ( 1 , v , 0 ) , k ′ = E ( 1 , cos ⁡ θ , − sin ⁡ θ ) , p ′ = m V e ′ (3.465) k = E 0 ( 1 , 1 , 0 ) , p = m γ ( 1 , v , 0 ) , k ′ = E ( 1 , cos ⁡ θ , − sin ⁡ θ ) , p ′ = m V e ′ {:(3.465)k=E_(0)(1","1","0)","quad p=m gamma(1","v","0)","quadk^(')=E(1","cos theta","-sin theta)","quadp^(')=mV_(e)^('):}\begin{equation*} k=E_{0}(1,1,0), \quad p=m \gamma(1, v, 0), \quad k^{\prime}=E(1, \cos \theta,-\sin \theta), \quad p^{\prime}=m V_{e}^{\prime} \tag{3.465} \end{equation*}(3.465)k=E0(1,1,0),p=mγ(1,v,0),k′=E(1,cos⁡θ,−sin⁡θ),p′=mVe′
where E 0 E 0 E_(0)E_{0}E0 is the initial photon energy, v v vvv is the initial speed of the electron, and E E EEE is the sought energy of the photon after scattering. By conservation of 4-momentum, we find that
(3.466) p + k − k ′ = p ′ ⇒ ( p + k − k ′ ) 2 = m 2 (3.466) p + k − k ′ = p ′ ⇒ p + k − k ′ 2 = m 2 {:(3.466)p+k-k^(')=p^(')quad=>quad(p+k-k^('))^(2)=m^(2):}\begin{equation*} p+k-k^{\prime}=p^{\prime} \quad \Rightarrow \quad\left(p+k-k^{\prime}\right)^{2}=m^{2} \tag{3.466} \end{equation*}(3.466)p+k−k′=p′⇒(p+k−k′)2=m2
This leads to
m 2 + 2 p ⋅ k − 2 ( p + k ) ⋅ k ′ = m 2 ⇒ p ⋅ k = ( p + k ) ⋅ k ′ (3.467) ⇒ m γ E 0 ( 1 − v ) = m γ E ( 1 − v cos ⁡ θ ) + E E 0 ( 1 − cos ⁡ θ ) m 2 + 2 p ⋅ k − 2 ( p + k ) ⋅ k ′ = m 2 ⇒ p ⋅ k = ( p + k ) ⋅ k ′ (3.467) ⇒ m γ E 0 ( 1 − v ) = m γ E ( 1 − v cos ⁡ θ ) + E E 0 ( 1 − cos ⁡ θ ) {:[m^(2)+2p*k-2(p+k)*k^(')=m^(2)quad=>quad p*k=(p+k)*k^(')],[(3.467)=>quad m gammaE_(0)(1-v)=m gamma E(1-v cos theta)+EE_(0)(1-cos theta)]:}\begin{align*} & m^{2}+2 p \cdot k-2(p+k) \cdot k^{\prime}=m^{2} \quad \Rightarrow \quad p \cdot k=(p+k) \cdot k^{\prime} \\ & \Rightarrow \quad m \gamma E_{0}(1-v)=m \gamma E(1-v \cos \theta)+E E_{0}(1-\cos \theta) \tag{3.467} \end{align*}m2+2p⋅k−2(p+k)⋅k′=m2⇒p⋅k=(p+k)⋅k′(3.467)⇒mγE0(1−v)=mγE(1−vcos⁡θ)+EE0(1−cos⁡θ)
Figure 3.8 Setup of inverse Compton scattering γ + e − ⟶ γ + e − γ + e − ⟶ γ + e − gamma+e^(-)longrightarrow gamma+e^(-)\gamma+e^{-} \longrightarrow \gamma+e^{-}γ+e−⟶γ+e−before and after scattering.
Solving for E E EEE leads to
(3.468) E = m γ E 0 ( 1 − v ) m γ ( 1 − v cos ⁡ θ ) + E 0 ( 1 − cos ⁡ θ ) (3.468) E = m γ E 0 ( 1 − v ) m γ ( 1 − v cos ⁡ θ ) + E 0 ( 1 − cos ⁡ θ ) {:(3.468)E=(m gammaE_(0)(1-v))/(m gamma(1-v cos theta)+E_(0)(1-cos theta)):}\begin{equation*} E=\frac{m \gamma E_{0}(1-v)}{m \gamma(1-v \cos \theta)+E_{0}(1-\cos \theta)} \tag{3.468} \end{equation*}(3.468)E=mγE0(1−v)mγ(1−vcos⁡θ)+E0(1−cos⁡θ)
1.99
Consider the reaction μ + ⟶ e + + v e + v ¯ μ μ + ⟶ e + + v e + v ¯ μ mu^(+)longrightarrowe^(+)+v_(e)+ bar(v)_(mu)\mu^{+} \longrightarrow e^{+}+v_{e}+\bar{v}_{\mu}μ+⟶e++ve+v¯μ. Let p μ , p e , p v e p μ , p e , p v e p_(mu),p_(e),p_(v_(e))p_{\mu}, p_{e}, p_{v_{e}}pμ,pe,pve, and p v ¯ μ p v ¯ μ p_( bar(v)_(mu))p_{\bar{v}_{\mu}}pv¯μ be the 4-momenta of the antimuon, positron, electron neutrino, and antimuon neutrino, respectively. Conservation of 4-momentum gives p μ = p e + p v e + p v ¯ μ p μ = p e + p v e + p v ¯ μ p_(mu)=p_(e)+p_(v_(e))+p_( bar(v)_(mu))p_{\mu}=p_{e}+p_{v_{e}}+p_{\bar{v}_{\mu}}pμ=pe+pve+pv¯μ, so that p μ − p v e = p e + p v ¯ μ ≡ P p μ − p v e = p e + p v ¯ μ ≡ P p_(mu)-p_(v_(e))=p_(e)+p_( bar(v)_(mu))-=Pp_{\mu}-p_{v_{e}}=p_{e}+p_{\bar{v}_{\mu}} \equiv Ppμ−pve=pe+pv¯μ≡P. Squaring both sides yields
P 2 = p μ 2 − 2 p μ ⋅ p v e + p v e 2 = m μ 2 − 2 p μ ⋅ p v e + m v e 2 (3.469) = ( p e + p v ¯ μ ) 2 = m e 2 + 2 p e ⋅ p v ¯ μ + m v ¯ μ 2 P 2 = p μ 2 − 2 p μ ⋅ p v e + p v e 2 = m μ 2 − 2 p μ ⋅ p v e + m v e 2 (3.469) = p e + p v ¯ μ 2 = m e 2 + 2 p e ⋅ p v ¯ μ + m v ¯ μ 2 {:[P^(2)=p_(mu)^(2)-2p_(mu)*p_(v_(e))+p_(v_(e))^(2)=m_(mu)^(2)-2p_(mu)*p_(v_(e))+m_(v_(e))^(2)],[(3.469)=(p_(e)+p_( bar(v)_(mu)))^(2)=m_(e)^(2)+2p_(e)*p_( bar(v)_(mu))+m_( bar(v)_(mu))^(2)]:}\begin{align*} P^{2}=p_{\mu}^{2}-2 p_{\mu} \cdot p_{v_{e}}+p_{v_{e}}^{2} & =m_{\mu}^{2}-2 p_{\mu} \cdot p_{v_{e}}+m_{v_{e}}^{2} \\ & =\left(p_{e}+p_{\bar{v}_{\mu}}\right)^{2}=m_{e}^{2}+2 p_{e} \cdot p_{\bar{v}_{\mu}}+m_{\bar{v}_{\mu}}^{2} \tag{3.469} \end{align*}P2=pμ2−2pμ⋅pve+pve2=mμ2−2pμ⋅pve+mve2(3.469)=(pe+pv¯μ)2=me2+2pe⋅pv¯μ+mv¯μ2
Neglecting the neutrino masses compared to lepton masses yields the relation
(3.470) m μ 2 − 2 p μ ⋅ p v e ≃ m e 2 + 2 p e ⋅ p v ¯ μ (3.470) m μ 2 − 2 p μ ⋅ p v e ≃ m e 2 + 2 p e ⋅ p v ¯ μ {:(3.470)m_(mu)^(2)-2p_(mu)*p_(v_(e))≃m_(e)^(2)+2p_(e)*p_( bar(v)_(mu)):}\begin{equation*} m_{\mu}^{2}-2 p_{\mu} \cdot p_{v_{e}} \simeq m_{e}^{2}+2 p_{e} \cdot p_{\bar{v}_{\mu}} \tag{3.470} \end{equation*}(3.470)mμ2−2pμ⋅pve≃me2+2pe⋅pv¯μ
All of the terms in this relation are Lorentz invariant and may be calculated in any inertial system. Calculating p e ⋅ p v ¯ μ p e ⋅ p v ¯ μ p_(e)*p_( bar(v)_(mu))p_{e} \cdot p_{\bar{v}_{\mu}}pe⋅pv¯μ in the rest frame of the positron, it is easy to obtain p e ⋅ p v ¯ μ ≥ m e m v ¯ μ p e ⋅ p v ¯ μ ≥ m e m v ¯ μ p_(e)*p_( bar(v)_(mu)) >= m_(e)m_( bar(v)_(mu))p_{e} \cdot p_{\bar{v}_{\mu}} \geq m_{e} m_{\bar{v}_{\mu}}pe⋅pv¯μ≥memv¯μ, which is negligible compared to m e 2 m e 2 m_(e)^(2)m_{e}^{2}me2. In the rest frame of the antimuon, we have p μ = ( m μ , 0 ) p μ = m μ , 0 p_(mu)=(m_(mu),0)p_{\mu}=\left(m_{\mu}, \mathbf{0}\right)pμ=(mμ,0) and p v e = ( E , p ) p v e = ( E , p ) p_(v_(e))=(E,p)p_{v_{e}}=(E, \boldsymbol{p})pve=(E,p), where | p | = E 2 − m v e 2 | p | = E 2 − m v e 2 |p|=sqrt(E^(2)-m_(v_(e))^(2))|\boldsymbol{p}|=\sqrt{E^{2}-m_{v_{e}}^{2}}|p|=E2−mve2 and E E EEE being the total energy of the electron neutrino. Inserting this into the equation above results in the inequality
(3.471) m μ 2 − 2 m μ E ≥ m e 2 ⇒ E ≤ m μ 2 − m e 2 2 m μ (3.471) m μ 2 − 2 m μ E ≥ m e 2 ⇒ E ≤ m μ 2 − m e 2 2 m μ {:(3.471)m_(mu)^(2)-2m_(mu)E >= m_(e)^(2)quad=>quad E <= (m_(mu)^(2)-m_(e)^(2))/(2m_(mu)):}\begin{equation*} m_{\mu}^{2}-2 m_{\mu} E \geq m_{e}^{2} \quad \Rightarrow \quad E \leq \frac{m_{\mu}^{2}-m_{e}^{2}}{2 m_{\mu}} \tag{3.471} \end{equation*}(3.471)mμ2−2mμE≥me2⇒E≤mμ2−me22mμ
Thus, the largest possible total energy of the electron neutrino in the rest frame of the antimuon is given by
(3.472) E max = m μ 2 − m e 2 2 m μ (3.472) E max = m μ 2 − m e 2 2 m μ {:(3.472)E_(max)=(m_(mu)^(2)-m_(e)^(2))/(2m_(mu)):}\begin{equation*} E_{\max }=\frac{m_{\mu}^{2}-m_{e}^{2}}{2 m_{\mu}} \tag{3.472} \end{equation*}(3.472)Emax=mμ2−me22mμ

1.100

Consider the decay ρ ⟶ μ − + μ + + γ ρ ⟶ μ − + μ + + γ rho longrightarrowmu^(-)+mu^(+)+gamma\rho \longrightarrow \mu^{-}+\mu^{+}+\gammaρ⟶μ−+μ++γ. Conservation of 4-momentum gives the relation
(3.473) p ρ − p μ + = p μ − + p γ (3.473) p ρ − p μ + = p μ − + p γ {:(3.473)p_(rho)-p_(mu^(+))=p_(mu^(-))+p_(gamma):}\begin{equation*} p_{\rho}-p_{\mu^{+}}=p_{\mu^{-}}+p_{\gamma} \tag{3.473} \end{equation*}(3.473)pρ−pμ+=pμ−+pγ
Squaring this expression yields
(3.474) ( p ρ − p μ + ) 2 = ( p μ − + p γ ) 2 (3.474) p ρ − p μ + 2 = p μ − + p γ 2 {:(3.474)(p_(rho)-p_(mu^(+)))^(2)=(p_(mu^(-))+p_(gamma))^(2):}\begin{equation*} \left(p_{\rho}-p_{\mu^{+}}\right)^{2}=\left(p_{\mu^{-}}+p_{\gamma}\right)^{2} \tag{3.474} \end{equation*}(3.474)(pρ−pμ+)2=(pμ−+pγ)2
where both sides are clearly Lorentz invariant. In the rest frame of the ρ ρ rho\rhoρ-meson, the left-hand side of this expression is given by
(3.475) ( p ρ − p μ + ) 2 = m ρ 2 + m μ 2 − 2 p ρ ⋅ p μ + = m ρ 2 + m μ 2 − 2 m ρ ( T μ + m μ ) (3.475) p ρ − p μ + 2 = m ρ 2 + m μ 2 − 2 p ρ ⋅ p μ + = m ρ 2 + m μ 2 − 2 m ρ T μ + m μ {:(3.475)(p_(rho)-p_(mu^(+)))^(2)=m_(rho)^(2)+m_(mu)^(2)-2p_(rho)*p_(mu^(+))=m_(rho)^(2)+m_(mu)^(2)-2m_(rho)(T_(mu)+m_(mu)):}\begin{equation*} \left(p_{\rho}-p_{\mu^{+}}\right)^{2}=m_{\rho}^{2}+m_{\mu}^{2}-2 p_{\rho} \cdot p_{\mu^{+}}=m_{\rho}^{2}+m_{\mu}^{2}-2 m_{\rho}\left(T_{\mu}+m_{\mu}\right) \tag{3.475} \end{equation*}(3.475)(pρ−pμ+)2=mρ2+mμ2−2pρ⋅pμ+=mρ2+mμ2−2mρ(Tμ+mμ)
where m μ ≡ m μ − = m μ + m μ ≡ m μ − = m μ + m_(mu)-=m_(mu^(-))=m_(mu^(+))m_{\mu} \equiv m_{\mu^{-}}=m_{\mu^{+}}mμ≡mμ−=mμ+and T μ T μ T_(mu)T_{\mu}Tμ is the kinetic energy of the μ + μ + mu^(+)\mu^{+}μ+. In the rest frame of the μ − μ − mu^(-)\mu^{-}μ−, the right-hand side becomes
(3.476) ( p μ − + p γ ) 2 = m μ ( m μ + 2 k ) ≥ m μ 2 (3.476) p μ − + p γ 2 = m μ m μ + 2 k ≥ m μ 2 {:(3.476)(p_(mu^(-))+p_(gamma))^(2)=m_(mu)(m_(mu)+2k) >= m_(mu)^(2):}\begin{equation*} \left(p_{\mu^{-}}+p_{\gamma}\right)^{2}=m_{\mu}\left(m_{\mu}+2 k\right) \geq m_{\mu}^{2} \tag{3.476} \end{equation*}(3.476)(pμ−+pγ)2=mμ(mμ+2k)≥mμ2
where k k kkk is the energy of the γ γ gamma\gammaγ. It follows that
(3.477) m ρ 2 + m μ 2 − 2 m ρ ( T μ + m μ ) ≥ m μ 2 ⇒ T μ ≤ m ρ 2 − m μ ≃ 279 MeV (3.477) m ρ 2 + m μ 2 − 2 m ρ T μ + m μ ≥ m μ 2 ⇒ T μ ≤ m ρ 2 − m μ ≃ 279 MeV {:(3.477)m_(rho)^(2)+m_(mu)^(2)-2m_(rho)(T_(mu)+m_(mu)) >= m_(mu)^(2)quad=>quadT_(mu) <= (m_(rho))/(2)-m_(mu)≃279MeV:}\begin{equation*} m_{\rho}^{2}+m_{\mu}^{2}-2 m_{\rho}\left(T_{\mu}+m_{\mu}\right) \geq m_{\mu}^{2} \quad \Rightarrow \quad T_{\mu} \leq \frac{m_{\rho}}{2}-m_{\mu} \simeq 279 \mathrm{MeV} \tag{3.477} \end{equation*}(3.477)mρ2+mμ2−2mρ(Tμ+mμ)≥mμ2⇒Tμ≤mρ2−mμ≃279MeV
Alternatively, one may realize that the maximal energy of the muons is given when the energy of the photon goes to zero. In that case, since the muons have identical masses, the total energy of the ρ ρ rho\rhoρ-meson will be evenly divided to the total energy of the muons. The kinetic energy of one of the muons is then given by T μ = E μ − m μ T μ = E μ − m μ T_(mu)=E_(mu)-m_(mu)T_{\mu}=E_{\mu}-m_{\mu}Tμ=Eμ−mμ, where E μ = m ρ / 2 E μ = m ρ / 2 E_(mu)=m_(rho)//2E_{\mu}=m_{\rho} / 2Eμ=mρ/2 is the total energy of one of the muons. Thus, the maximal kinetic energy that one of the muons can have in this decay in the rest frame of the ρ ρ rho\rhoρ-meson is T μ = m ρ / 2 − m μ T μ = m ρ / 2 − m μ T_(mu)=m_(rho)//2-m_(mu)T_{\mu}=m_{\rho} / 2-m_{\mu}Tμ=mρ/2−mμ.

1.101

Giving all quantities in the rest frame of the 76 Ge 76 Ge ^(76)Ge{ }^{76} \mathrm{Ge}76Ge, we have that
(3.478) p Ge = ( m Ge , 0 ) , (3.479) p Se = ( E Se , p Se ) , (3.480) p e 1 = ( E 1 , p 1 ) , (3.481) p e 2 = ( E 2 , p 2 ) (3.478) p Ge = m Ge , 0 , (3.479) p Se = E Se , p Se , (3.480) p e 1 = E 1 , p 1 , (3.481) p e 2 = E 2 , p 2 {:[(3.478)p_(Ge)=(m_(Ge),0)","],[(3.479)p_(Se)=(E_(Se),p_(Se))","],[(3.480)p_(e1)=(E_(1),p_(1))","],[(3.481)p_(e2)=(E_(2),p_(2))]:}\begin{align*} p_{\mathrm{Ge}} & =\left(m_{\mathrm{Ge}}, \mathbf{0}\right), \tag{3.478}\\ p_{\mathrm{Se}} & =\left(E_{\mathrm{Se}}, \boldsymbol{p}_{\mathrm{Se}}\right), \tag{3.479}\\ p_{e 1} & =\left(E_{1}, \boldsymbol{p}_{1}\right), \tag{3.480}\\ p_{e 2} & =\left(E_{2}, \boldsymbol{p}_{2}\right) \tag{3.481} \end{align*}(3.478)pGe=(mGe,0),(3.479)pSe=(ESe,pSe),(3.480)pe1=(E1,p1),(3.481)pe2=(E2,p2)
The conservation of 4-momentum states that
(3.482) p Ge = p Se + p e 1 + p e 2 (3.482) p Ge = p Se + p e 1 + p e 2 {:(3.482)p_(Ge)=p_(Se)+p_(e1)+p_(e2):}\begin{equation*} p_{\mathrm{Ge}}=p_{\mathrm{Se}}+p_{e 1}+p_{e 2} \tag{3.482} \end{equation*}(3.482)pGe=pSe+pe1+pe2
In particular, the time component of this relation states that
(3.483) m Ge − E Se = E 1 + E 2 = E = T + 2 m e (3.483) m Ge − E Se = E 1 + E 2 = E = T + 2 m e {:(3.483)m_(Ge)-E_(Se)=E_(1)+E_(2)=E=T+2m_(e):}\begin{equation*} m_{\mathrm{Ge}}-E_{\mathrm{Se}}=E_{1}+E_{2}=E=T+2 m_{e} \tag{3.483} \end{equation*}(3.483)mGe−ESe=E1+E2=E=T+2me
where E E EEE is the total energy of the two electrons and T T TTT is their total kinetic energy, the quantity in which we are interested. Since the masses are invariant, we deduce that
(3.484) T = m Ge − E Se − 2 m e (3.484) T = m Ge − E Se − 2 m e {:(3.484)T=m_(Ge)-E_(Se)-2m_(e):}\begin{equation*} T=m_{\mathrm{Ge}}-E_{\mathrm{Se}}-2 m_{e} \tag{3.484} \end{equation*}(3.484)T=mGe−ESe−2me
is a function of E Se E Se E_(Se)E_{\mathrm{Se}}ESe only. In order to maximize T , E Se = m Se 2 + p Se 2 T , E Se = m Se 2 + p Se 2 T,E_(Se)=sqrt(m_(Se)^(2)+p_(Se)^(2))T, E_{\mathrm{Se}}=\sqrt{m_{\mathrm{Se}}^{2}+p_{\mathrm{Se}}^{2}}T,ESe=mSe2+pSe2 must take its minimum allowed value, i.e., E Se = m Se E Se = m Se E_(Se)=m_(Se)E_{\mathrm{Se}}=m_{\mathrm{Se}}ESe=mSe. It is necessary to check that this is kinematically allowed, which is the case since the momentum conservation is
solved by p 1 = − p 2 p 1 = − p 2 p_(1)=-p_(2)\boldsymbol{p}_{1}=-\boldsymbol{p}_{2}p1=−p2, which leaves the energy of the electrons as a free parameter, which may be adjusted to solve the energy conservation. It follows that
(3.485) T ≤ m Ge − m Se − 2 m e ≡ T max (3.485) T ≤ m Ge − m Se − 2 m e ≡ T max {:(3.485)T <= m_(Ge)-m_(Se)-2m_(e)-=T_(max):}\begin{equation*} T \leq m_{\mathrm{Ge}}-m_{\mathrm{Se}}-2 m_{e} \equiv T_{\max } \tag{3.485} \end{equation*}(3.485)T≤mGe−mSe−2me≡Tmax
The minimal energy of the electrons is instead obtained when the 76 Se 76 Se ^(76)Se{ }^{76} \mathrm{Se}76Se obtains its maximal energy. Conservation of 4-momentum gives
(3.486) ( p Ge − p Se ) 2 = ( p e 1 + p e 2 ) 2 ≥ 4 m e 2 (3.486) p Ge − p Se 2 = p e 1 + p e 2 2 ≥ 4 m e 2 {:(3.486)(p_(Ge)-p_(Se))^(2)=(p_(e1)+p_(e2))^(2) >= 4m_(e)^(2):}\begin{equation*} \left(p_{\mathrm{Ge}}-p_{\mathrm{Se}}\right)^{2}=\left(p_{e 1}+p_{e 2}\right)^{2} \geq 4 m_{e}^{2} \tag{3.486} \end{equation*}(3.486)(pGe−pSe)2=(pe1+pe2)2≥4me2
The left-hand side of this equation evaluates to
(3.487) ( p Ge − p Se ) 2 = m Ge 2 + m Se 2 − 2 m Ge E Se (3.487) p Ge − p Se 2 = m Ge 2 + m Se 2 − 2 m Ge E Se {:(3.487)(p_(Ge)-p_(Se))^(2)=m_(Ge)^(2)+m_(Se)^(2)-2m_(Ge)E_(Se):}\begin{equation*} \left(p_{\mathrm{Ge}}-p_{\mathrm{Se}}\right)^{2}=m_{\mathrm{Ge}}^{2}+m_{\mathrm{Se}}^{2}-2 m_{\mathrm{Ge}} E_{\mathrm{Se}} \tag{3.487} \end{equation*}(3.487)(pGe−pSe)2=mGe2+mSe2−2mGeESe
Solving for E Se E Se E_(Se)E_{\mathrm{Se}}ESe, we obtain
(3.488) E Se ≤ m Ge 2 + m Se 2 − 4 m e 2 2 m Ge (3.488) E Se ≤ m Ge 2 + m Se 2 − 4 m e 2 2 m Ge {:(3.488)E_(Se) <= (m_(Ge)^(2)+m_(Se)^(2)-4m_(e)^(2))/(2m_(Ge)):}\begin{equation*} E_{\mathrm{Se}} \leq \frac{m_{\mathrm{Ge}}^{2}+m_{\mathrm{Se}}^{2}-4 m_{e}^{2}}{2 m_{\mathrm{Ge}}} \tag{3.488} \end{equation*}(3.488)ESe≤mGe2+mSe2−4me22mGe
and thus, we have
(3.489) T ≥ ( m Ge − 2 m e ) 2 − m Se 2 2 m Ge ≡ T min (3.489) T ≥ m Ge − 2 m e 2 − m Se 2 2 m Ge ≡ T min {:(3.489)T >= ((m_(Ge)-2m_(e))^(2)-m_(Se)^(2))/(2m_(Ge))-=T_(min):}\begin{equation*} T \geq \frac{\left(m_{\mathrm{Ge}}-2 m_{e}\right)^{2}-m_{\mathrm{Se}}^{2}}{2 m_{\mathrm{Ge}}} \equiv T_{\mathrm{min}} \tag{3.489} \end{equation*}(3.489)T≥(mGe−2me)2−mSe22mGe≡Tmin
It is of interest to note that
(3.490) T max − T min T max = m Ge − m Se + 2 m e 2 m Ge (3.490) T max − T min T max = m Ge − m Se + 2 m e 2 m Ge {:(3.490)(T_(max)-T_(min))/(T_(max))=(m_(Ge)-m_(Se)+2m_(e))/(2m_(Ge)):}\begin{equation*} \frac{T_{\max }-T_{\min }}{T_{\max }}=\frac{m_{\mathrm{Ge}}-m_{\mathrm{Se}}+2 m_{e}}{2 m_{\mathrm{Ge}}} \tag{3.490} \end{equation*}(3.490)Tmax−TminTmax=mGe−mSe+2me2mGe
which is typically a very small number since the the difference between the masses of the nuclei is relatively small. Thus, the electron spectrum for the neutrinoless double beta decay is very peaked. This is in sharp contrast to the case of the more common double beta decay ( X ⟶ Y + 2 e − + 2 v ¯ e X ⟶ Y + 2 e − + 2 v ¯ e X longrightarrow Y+2e^(-)+2 bar(v)_(e)X \longrightarrow Y+2 e^{-}+2 \bar{v}_{e}X⟶Y+2e−+2v¯e ), where the electron spectrum is continuous and broad due to the possibility of the neutrinos taking some of the energy.

1.102

According to the conservation of 4-momentum, the 4-momentum of the new particle ϕ ϕ phi\phiϕ must be given by
(3.491) p ϕ = p 1 + p 2 (3.491) p ϕ = p 1 + p 2 {:(3.491)p_(phi)=p_(1)+p_(2):}\begin{equation*} p_{\phi}=p_{1}+p_{2} \tag{3.491} \end{equation*}(3.491)pϕ=p1+p2
In general, the magnitude of a particle's 4-momentum is its mass, and therefore, we find that
(3.492) m ϕ 2 = p ϕ 2 = ( p 1 + p 2 ) 2 = 2 p 1 ⋅ p 2 (3.492) m ϕ 2 = p ϕ 2 = p 1 + p 2 2 = 2 p 1 ⋅ p 2 {:(3.492)m_(phi)^(2)=p_(phi)^(2)=(p_(1)+p_(2))^(2)=2p_(1)*p_(2):}\begin{equation*} m_{\phi}^{2}=p_{\phi}^{2}=\left(p_{1}+p_{2}\right)^{2}=2 p_{1} \cdot p_{2} \tag{3.492} \end{equation*}(3.492)mϕ2=pϕ2=(p1+p2)2=2p1⋅p2
since p 1 2 = p 2 2 = 0 p 1 2 = p 2 2 = 0 p_(1)^(2)=p_(2)^(2)=0p_{1}^{2}=p_{2}^{2}=0p12=p22=0. Computing the remaining inner product in the lab frame, we find that
(3.493) m ϕ 2 = 2 ω 1 ω 2 ( 1 − cos ⁡ θ ) (3.493) m ϕ 2 = 2 ω 1 ω 2 ( 1 − cos ⁡ θ ) {:(3.493)m_(phi)^(2)=2omega_(1)omega_(2)(1-cos theta):}\begin{equation*} m_{\phi}^{2}=2 \omega_{1} \omega_{2}(1-\cos \theta) \tag{3.493} \end{equation*}(3.493)mϕ2=2ω1ω2(1−cos⁡θ)

1.103

The total 4-momentum is:
  1. In the laboratory (lab) system: p lab = ( E + m p , p , 0 , 0 ) p lab  = E + m p , p , 0 , 0 p_("lab ")=(E+m_(p),p,0,0)p_{\text {lab }}=\left(E+m_{p}, p, 0,0\right)plab =(E+mp,p,0,0),
  2. In the center-of-mass (CM) system: p CM = ( E ∗ , 0 , 0 , 0 ) p CM = E ∗ , 0 , 0 , 0 p_(CM)=(E_(**),0,0,0)p_{\mathrm{CM}}=\left(E_{*}, 0,0,0\right)pCM=(E∗,0,0,0),
    where E E EEE and p p ppp are the energy and the momentum of the incoming proton, respectively, E ∗ E ∗ E_(**)E_{*}E∗ is the energy in the CM system, and m p m p m_(p)m_{p}mp is the rest energy (mass) for a proton (or antiproton).
The 4-momentum squared is an invariant, and thus, the same in the two systems. Therefore, we have
(3.494) p lab 2 = p CM 2 (3.494) p lab 2 = p CM 2 {:(3.494)p_(lab)^(2)=p_(CM)^(2):}\begin{equation*} p_{\mathrm{lab}}^{2}=p_{\mathrm{CM}}^{2} \tag{3.494} \end{equation*}(3.494)plab2=pCM2
which gives
(3.495) ( E + m p ) 2 − p 2 = E ∗ 2 (3.495) E + m p 2 − p 2 = E ∗ 2 {:(3.495)(E+m_(p))^(2)-p^(2)=E_(**)^(2):}\begin{equation*} \left(E+m_{p}\right)^{2}-p^{2}=E_{*}^{2} \tag{3.495} \end{equation*}(3.495)(E+mp)2−p2=E∗2
Inserting E 2 = m p 2 + p 2 E 2 = m p 2 + p 2 E^(2)=m_(p)^(2)+p^(2)E^{2}=m_{p}^{2}+p^{2}E2=mp2+p2 and T = E − m p T = E − m p T=E-m_(p)T=E-m_{p}T=E−mp, i.e., the kinetic energy of the incoming proton, we obtain
(3.496) E ∗ = 2 m p ( T + 2 m p ) (3.496) E ∗ = 2 m p T + 2 m p {:(3.496)E_(**)=sqrt(2m_(p)(T+2m_(p))):}\begin{equation*} E_{*}=\sqrt{2 m_{p}\left(T+2 m_{p}\right)} \tag{3.496} \end{equation*}(3.496)E∗=2mp(T+2mp)
A necessary condition for production of a proton-antiproton pair is E ∗ ≥ 4 m p E ∗ ≥ 4 m p E_(**) >= 4m_(p)E_{*} \geq 4 m_{p}E∗≥4mp, i.e.,
(3.497) T ≥ 6 m p ≃ 6.938 MeV ≈ 5628 MeV (3.497) T ≥ 6 m p ≃ 6.938 MeV ≈ 5628 MeV {:(3.497)T >= 6m_(p)≃6.938MeV~~5628MeV:}\begin{equation*} T \geq 6 m_{p} \simeq 6.938 \mathrm{MeV} \approx 5628 \mathrm{MeV} \tag{3.497} \end{equation*}(3.497)T≥6mp≃6.938MeV≈5628MeV
The kinetic energy 8000 MeV is therefore sufficient.

1.104

The same method as in Problem 1.103 gives that the kinetic energy of the negative pion must satisfy the inequality
(3.498) T π > 1 2 m p ( 3 m π 2 + 4 m π m n + m n 2 − m p 2 ) − m π ≈ 174 MeV (3.498) T π > 1 2 m p 3 m π 2 + 4 m π m n + m n 2 − m p 2 − m π ≈ 174 MeV {:(3.498)T_(pi) > (1)/(2m_(p))(3m_(pi)^(2)+4m_(pi)m_(n)+m_(n)^(2)-m_(p)^(2))-m_(pi)~~174MeV:}\begin{equation*} T_{\pi}>\frac{1}{2 m_{p}}\left(3 m_{\pi}^{2}+4 m_{\pi} m_{n}+m_{n}^{2}-m_{p}^{2}\right)-m_{\pi} \approx 174 \mathrm{MeV} \tag{3.498} \end{equation*}(3.498)Tπ>12mp(3mπ2+4mπmn+mn2−mp2)−mπ≈174MeV

1.105

a) Before the reaction, we have
(3.499) P in 2 = ( p p + p d ) 2 = ( m p + m d ) 2 + 2 T p m d ≃ 9 m 2 + 4 m T p (3.499) P in 2 = p p + p d 2 = m p + m d 2 + 2 T p m d ≃ 9 m 2 + 4 m T p {:(3.499)P_(in)^(2)=(p_(p)+p_(d))^(2)=(m_(p)+m_(d))^(2)+2T_(p)m_(d)≃9m^(2)+4mT_(p):}\begin{equation*} P_{\mathrm{in}}^{2}=\left(p_{p}+p_{d}\right)^{2}=\left(m_{p}+m_{d}\right)^{2}+2 T_{p} m_{d} \simeq 9 m^{2}+4 m T_{p} \tag{3.499} \end{equation*}(3.499)Pin2=(pp+pd)2=(mp+md)2+2Tpmd≃9m2+4mTp
where m m mmm is the nucleon mass. On the other hand, after the reaction, we have
P out 2 = ( p p + p p ′ + p n + p η ) 2 ≥ ( m p + m p + m n + m η ) 2 ≃ ( 3 m + m η ) 2 (3.500) = 9 m 2 + 6 m m η + m η 2 . P out  2 = p p + p p ′ + p n + p η 2 ≥ m p + m p + m n + m η 2 ≃ 3 m + m η 2 (3.500) = 9 m 2 + 6 m m η + m η 2 . {:[P_("out ")^(2)=(p_(p)+p_(p^('))+p_(n)+p_(eta))^(2) >= (m_(p)+m_(p)+m_(n)+m_(eta))^(2)≃(3m+m_(eta))^(2)],[(3.500)=9m^(2)+6mm_(eta)+m_(eta)^(2).]:}\begin{align*} P_{\text {out }}^{2} & =\left(p_{p}+p_{p^{\prime}}+p_{n}+p_{\eta}\right)^{2} \geq\left(m_{p}+m_{p}+m_{n}+m_{\eta}\right)^{2} \simeq\left(3 m+m_{\eta}\right)^{2} \\ & =9 m^{2}+6 m m_{\eta}+m_{\eta}^{2} . \tag{3.500} \end{align*}Pout 2=(pp+pp′+pn+pη)2≥(mp+mp+mn+mη)2≃(3m+mη)2(3.500)=9m2+6mmη+mη2.
Using P in 2 = P out 2 P in 2 = P out  2 P_(in)^(2)=P_("out ")^(2)P_{\mathrm{in}}^{2}=P_{\text {out }}^{2}Pin2=Pout 2, this gives
(3.501) T p ≃ m η ( 6 m + m η ) 4 m ≥ 13 8 m η ≈ 900 MeV > 700 MeV (3.501) T p ≃ m η 6 m + m η 4 m ≥ 13 8 m η ≈ 900 MeV > 700 MeV {:(3.501)T_(p)≃(m_(eta)(6m+m_(eta)))/(4m) >= (13)/(8)m_(eta)~~900MeV > 700MeV:}\begin{equation*} T_{p} \simeq \frac{m_{\eta}\left(6 m+m_{\eta}\right)}{4 m} \geq \frac{13}{8} m_{\eta} \approx 900 \mathrm{MeV}>700 \mathrm{MeV} \tag{3.501} \end{equation*}(3.501)Tp≃mη(6m+mη)4m≥138mη≈900MeV>700MeV
The reaction is therefore not possible at this kinetic energy.
b) Let the 4-momentum of the η η eta\etaη be p η = ( m η + T , p η ) p η = m η + T , p η p_(eta)=(m_(eta)+T,p_(eta))p_{\eta}=\left(m_{\eta}+T, \boldsymbol{p}_{\eta}\right)pη=(mη+T,pη), where T T TTT is the kinetic energy and p η p η p_(eta)\boldsymbol{p}_{\eta}pη is the 3-momentum in the rest frame of the nucleons. Squaring the relation from conservation of 4 -momentum now results in
(3.502) 9 m 2 + 4 m T p = [ ( 3 m , 0 ) + p η ] 2 = 9 m 2 + 6 m ( m η + T ) + m η 2 (3.502) 9 m 2 + 4 m T p = ( 3 m , 0 ) + p η 2 = 9 m 2 + 6 m m η + T + m η 2 {:(3.502)9m^(2)+4mT_(p)=[(3m,0)+p_(eta)]^(2)=9m^(2)+6m(m_(eta)+T)+m_(eta)^(2):}\begin{equation*} 9 m^{2}+4 m T_{p}=\left[(3 m, 0)+p_{\eta}\right]^{2}=9 m^{2}+6 m\left(m_{\eta}+T\right)+m_{\eta}^{2} \tag{3.502} \end{equation*}(3.502)9m2+4mTp=[(3m,0)+pη]2=9m2+6m(mη+T)+mη2
Solving for T T TTT results in
(3.503) T = 2 T p 3 − m η − m η 2 6 m ≃ 300 MeV (3.503) T = 2 T p 3 − m η − m η 2 6 m ≃ 300 MeV {:(3.503)T=(2T_(p))/(3)-m_(eta)-(m_(eta)^(2))/(6m)≃300MeV:}\begin{equation*} T=\frac{2 T_{p}}{3}-m_{\eta}-\frac{m_{\eta}^{2}}{6 m} \simeq 300 \mathrm{MeV} \tag{3.503} \end{equation*}(3.503)T=2Tp3−mη−mη26m≃300MeV

1.106

In general, we have the relation
(3.504) ( ∑ i p i ) 2 ≥ ( ∑ i m i ) 2 (3.504) ∑ i   p i 2 ≥ ∑ i   m i 2 {:(3.504)(sum_(i)p_(i))^(2) >= (sum_(i)m_(i))^(2):}\begin{equation*} \left(\sum_{i} p_{i}\right)^{2} \geq\left(\sum_{i} m_{i}\right)^{2} \tag{3.504} \end{equation*}(3.504)(∑ipi)2≥(∑imi)2
for an arbitrary sum of particle 4-momenta p i p i p_(i)p_{i}pi. For our two reactions, the conservation of 4-momentum yields
(3.505) p v + p X = p μ + p Y [ + p π ] (3.505) p v + p X = p μ + p Y + p π {:(3.505)p_(v)+p_(X)=p_(mu)+p_(Y)[+p_(pi)]:}\begin{equation*} p_{v}+p_{X}=p_{\mu}+p_{Y}\left[+p_{\pi}\right] \tag{3.505} \end{equation*}(3.505)pv+pX=pμ+pY[+pπ]
where the term in brackets only appears in the 1 π 1 π 1pi1 \pi1π reaction. Squaring this relation gives us
p v 2 + 2 p v ⋅ p X + p X 2 = 2 p v ⋅ p X + m X 2 = ( p μ + p Y [ + p π ] ) 2 ≥ ( m μ + m Y [ + m π ] ) 2 p v 2 + 2 p v ⋅ p X + p X 2 = 2 p v ⋅ p X + m X 2 = p μ + p Y + p π 2 ≥ m μ + m Y + m π 2 p_(v)^(2)+2p_(v)*p_(X)+p_(X)^(2)=2p_(v)*p_(X)+m_(X)^(2)=(p_(mu)+p_(Y)[+p_(pi)])^(2) >= (m_(mu)+m_(Y)[+m_(pi)])^(2)p_{v}^{2}+2 p_{v} \cdot p_{X}+p_{X}^{2}=2 p_{v} \cdot p_{X}+m_{X}^{2}=\left(p_{\mu}+p_{Y}\left[+p_{\pi}\right]\right)^{2} \geq\left(m_{\mu}+m_{Y}\left[+m_{\pi}\right]\right)^{2}pv2+2pv⋅pX+pX2=2pv⋅pX+mX2=(pμ+pY[+pπ])2≥(mμ+mY[+mπ])2.
In the rest frame of X X XXX, the product p v ⋅ p X p v ⋅ p X p_(v)*p_(X)p_{v} \cdot p_{X}pv⋅pX evaluates to
(3.507) p v ⋅ p X = E v m X (3.507) p v ⋅ p X = E v m X {:(3.507)p_(v)*p_(X)=E_(v)m_(X):}\begin{equation*} p_{v} \cdot p_{X}=E_{v} m_{X} \tag{3.507} \end{equation*}(3.507)pv⋅pX=EvmX
and we therefore obtain
(3.508) E v ≥ 1 2 m X { ( m μ + m Y [ + m π ] ) 2 − m X 2 } (3.508) E v ≥ 1 2 m X m μ + m Y + m π 2 − m X 2 {:(3.508)E_(v) >= (1)/(2m_(X)){(m_(mu)+m_(Y)[+m_(pi)])^(2)-m_(X)^(2)}:}\begin{equation*} E_{v} \geq \frac{1}{2 m_{X}}\left\{\left(m_{\mu}+m_{Y}\left[+m_{\pi}\right]\right)^{2}-m_{X}^{2}\right\} \tag{3.508} \end{equation*}(3.508)Ev≥12mX{(mμ+mY[+mπ])2−mX2}
where the lower limit is the threshold energy for the reaction to be possible. Dividing the two threshold energies, we obtain the ratio
(3.509) E v , th , 1 π E v , th, QE = ( m μ + m Y + m π ) 2 − m X 2 ( m μ + m Y ) 2 − m X 2 (3.509) E v ,  th  , 1 π E v ,  th,  QE = m μ + m Y + m π 2 − m X 2 m μ + m Y 2 − m X 2 {:(3.509)(E_(v," th ",1pi))/(E_(v," th, "QE))=((m_(mu)+m_(Y)+m_(pi))^(2)-m_(X)^(2))/((m_(mu)+m_(Y))^(2)-m_(X)^(2)):}\begin{equation*} \frac{E_{v, \text { th }, 1 \pi}}{E_{v, \text { th, } \mathrm{QE}}}=\frac{\left(m_{\mu}+m_{Y}+m_{\pi}\right)^{2}-m_{X}^{2}}{\left(m_{\mu}+m_{Y}\right)^{2}-m_{X}^{2}} \tag{3.509} \end{equation*}(3.509)Ev, th ,1πEv, th, QE=(mμ+mY+mπ)2−mX2(mμ+mY)2−mX2

1.107

We denote the incoming momenta p 1 p 1 p_(1)p_{1}p1 and p 2 p 2 p_(2)p_{2}p2, respectively, while we call the outgoing momenta k i , i = 1 , 2 , 3 , 4 k i , i = 1 , 2 , 3 , 4 k_(i),i=1,2,3,4k_{i}, i=1,2,3,4ki,i=1,2,3,4. Since all of the particles have the same mass, we have the relation p i 2 = k i 2 = m 2 p i 2 = k i 2 = m 2 p_(i)^(2)=k_(i)^(2)=m^(2)p_{i}^{2}=k_{i}^{2}=m^{2}pi2=ki2=m2. By conservation of 4-momentum, we have
(3.510) p 1 + p 2 = k 1 + k 2 + k 3 + k 4 (3.510) p 1 + p 2 = k 1 + k 2 + k 3 + k 4 {:(3.510)p_(1)+p_(2)=k_(1)+k_(2)+k_(3)+k_(4):}\begin{equation*} p_{1}+p_{2}=k_{1}+k_{2}+k_{3}+k_{4} \tag{3.510} \end{equation*}(3.510)p1+p2=k1+k2+k3+k4
Squaring this relation and using the inequality A ⋅ B ≥ A 2 B 2 A ⋅ B ≥ A 2 B 2 A*B >= sqrt(A^(2)B^(2))A \cdot B \geq \sqrt{A^{2} B^{2}}A⋅B≥A2B2 for any timelike vectors A A AAA and B B BBB, we find that
(3.511) 2 m 2 + 2 p 1 ⋅ p 2 ≥ ( 4 m ) 2 (3.511) 2 m 2 + 2 p 1 ⋅ p 2 ≥ ( 4 m ) 2 {:(3.511)2m^(2)+2p_(1)*p_(2) >= (4m)^(2):}\begin{equation*} 2 m^{2}+2 p_{1} \cdot p_{2} \geq(4 m)^{2} \tag{3.511} \end{equation*}(3.511)2m2+2p1⋅p2≥(4m)2
where the equality holds only if all of the outgoing particles are at relative rest, i.e., at the reaction threshold. At threshold, we therefore have
(3.512) p 1 ⋅ p 2 = 7 m 2 (3.512) p 1 ⋅ p 2 = 7 m 2 {:(3.512)p_(1)*p_(2)=7m^(2):}\begin{equation*} p_{1} \cdot p_{2}=7 m^{2} \tag{3.512} \end{equation*}(3.512)p1⋅p2=7m2
In the rest frame of one of the initial electrons (say the one with 4-momentum p 1 p 1 p_(1)p_{1}p1 ), we can write down the 4-momenta as
(3.513) p 1 = m ( 1 , 0 ) , p 2 = ( E , p ) (3.513) p 1 = m ( 1 , 0 ) , p 2 = ( E , p ) {:(3.513)p_(1)=m(1","0)","quadp_(2)=(E","p):}\begin{equation*} p_{1}=m(1, \mathbf{0}), \quad p_{2}=(E, \boldsymbol{p}) \tag{3.513} \end{equation*}(3.513)p1=m(1,0),p2=(E,p)
where E E EEE is the total energy of the other electron. It follows that
(3.514) m E = 7 m 2 ⇒ E = 7 m (3.514) m E = 7 m 2 ⇒ E = 7 m {:(3.514)mE=7m^(2)=>E=7m:}\begin{equation*} m E=7 m^{2} \Rightarrow E=7 m \tag{3.514} \end{equation*}(3.514)mE=7m2⇒E=7m
On the other hand, in the center-of-momentum frame, the total energy squared is the square of the total 4-momentum. Thus, the total center-of-mass (CM) energy is given by
(3.515) E CM = ( 4 m ) 2 = 4 m (3.515) E CM = ( 4 m ) 2 = 4 m {:(3.515)E_(CM)=sqrt((4m)^(2))=4m:}\begin{equation*} E_{\mathrm{CM}}=\sqrt{(4 m)^{2}}=4 m \tag{3.515} \end{equation*}(3.515)ECM=(4m)2=4m
at threshold. The ratio E / E CM E / E CM E//E_(CM)E / E_{\mathrm{CM}}E/ECM is therefore 7 / 4 7 / 4 7//47 / 47/4. Note that the total energy in the frame where one of the electrons is at rest is E + m = 8 m E + m = 8 m E+m=8mE+m=8 mE+m=8m.

1.108

In general, conservation of 4-momentum gives
(3.516) p 1 + p 2 = p 1 ′ + p 2 ′ + p K ′ (3.516) p 1 + p 2 = p 1 ′ + p 2 ′ + p K ′ {:(3.516)p_(1)+p_(2)=p_(1)^(')+p_(2)^(')+p_(K)^('):}\begin{equation*} p_{1}+p_{2}=p_{1}^{\prime}+p_{2}^{\prime}+p_{K}^{\prime} \tag{3.516} \end{equation*}(3.516)p1+p2=p1′+p2′+pK′
where p i p i p_(i)p_{i}pi are the 4-momenta of the incoming protons, p i ′ p i ′ p_(i)^(')p_{i}^{\prime}pi′ are the 4 -momenta of the outgoing protons, and p K ′ p K ′ p_(K)^(')p_{K}^{\prime}pK′ is the 4 -momentum of the kaon. By squaring this relation, we obtain
(3.517) p 1 2 + p 2 2 + 2 p 1 ⋅ p 2 = 2 m p 2 + 2 p 1 ⋅ p 2 = ( p 1 ′ + p 2 ′ + p K ′ ) 2 ≥ ( 2 m p + m K ) 2 (3.517) p 1 2 + p 2 2 + 2 p 1 ⋅ p 2 = 2 m p 2 + 2 p 1 ⋅ p 2 = p 1 ′ + p 2 ′ + p K ′ 2 ≥ 2 m p + m K 2 {:(3.517)p_(1)^(2)+p_(2)^(2)+2p_(1)*p_(2)=2m_(p)^(2)+2p_(1)*p_(2)=(p_(1)^(')+p_(2)^(')+p_(K)^('))^(2) >= (2m_(p)+m_(K))^(2):}\begin{equation*} p_{1}^{2}+p_{2}^{2}+2 p_{1} \cdot p_{2}=2 m_{p}^{2}+2 p_{1} \cdot p_{2}=\left(p_{1}^{\prime}+p_{2}^{\prime}+p_{K}^{\prime}\right)^{2} \geq\left(2 m_{p}+m_{K}\right)^{2} \tag{3.517} \end{equation*}(3.517)p12+p22+2p1⋅p2=2mp2+2p1⋅p2=(p1′+p2′+pK′)2≥(2mp+mK)2
where the equality holds at the threshold energy. In the first case where one of the protons is at rest, we find that
(3.518) p 1 = ( m p , 0 ) , p 2 = ( T + m p , p ) , (3.518) p 1 = m p , 0 , p 2 = T + m p , p , {:(3.518)p_(1)=(m_(p),0)","quadp_(2)=(T+m_(p),p)",":}\begin{equation*} p_{1}=\left(m_{p}, \mathbf{0}\right), \quad p_{2}=\left(T+m_{p}, \boldsymbol{p}\right), \tag{3.518} \end{equation*}(3.518)p1=(mp,0),p2=(T+mp,p),
where T T TTT is the kinetic energy of the moving proton and p p ppp is its momentum. This results in
(3.519) p 1 ⋅ p 2 = m p 2 + T m p (3.519) p 1 ⋅ p 2 = m p 2 + T m p {:(3.519)p_(1)*p_(2)=m_(p)^(2)+Tm_(p):}\begin{equation*} p_{1} \cdot p_{2}=m_{p}^{2}+T m_{p} \tag{3.519} \end{equation*}(3.519)p1⋅p2=mp2+Tmp
Inserting this into the relation above, we find that
(3.520) T th = m K 2 2 m p + 2 m K (3.520) T th = m K 2 2 m p + 2 m K {:(3.520)T_(th)=(m_(K)^(2))/(2m_(p))+2m_(K):}\begin{equation*} T_{\mathrm{th}}=\frac{m_{K}^{2}}{2 m_{p}}+2 m_{K} \tag{3.520} \end{equation*}(3.520)Tth=mK22mp+2mK
On the other hand, in the second case where both protons are moving with the same velocity, we have
(3.521) p 1 = ( T + m p , p ) , p 2 = ( T + m p , − p ) (3.521) p 1 = T + m p , p , p 2 = T + m p , − p {:(3.521)p_(1)=(T+m_(p),p)","quadp_(2)=(T+m_(p),-p):}\begin{equation*} p_{1}=\left(T+m_{p}, \boldsymbol{p}\right), \quad p_{2}=\left(T+m_{p},-\boldsymbol{p}\right) \tag{3.521} \end{equation*}(3.521)p1=(T+mp,p),p2=(T+mp,−p)
where T T TTT is the kinetic energy of one proton. As a result, we find
(3.522) p 1 + p 2 = 2 ( T + m p , 0 ) ⟹ ( p 1 + p 2 ) 2 = ( 2 T + 2 m p ) 2 ≥ ( 2 m p + m K ) 2 (3.522) p 1 + p 2 = 2 T + m p , 0 ⟹ p 1 + p 2 2 = 2 T + 2 m p 2 ≥ 2 m p + m K 2 {:(3.522)p_(1)+p_(2)=2(T+m_(p),0)quad Longrightarrowquad(p_(1)+p_(2))^(2)=(2T+2m_(p))^(2) >= (2m_(p)+m_(K))^(2):}\begin{equation*} p_{1}+p_{2}=2\left(T+m_{p}, \mathbf{0}\right) \quad \Longrightarrow \quad\left(p_{1}+p_{2}\right)^{2}=\left(2 T+2 m_{p}\right)^{2} \geq\left(2 m_{p}+m_{K}\right)^{2} \tag{3.522} \end{equation*}(3.522)p1+p2=2(T+mp,0)⟹(p1+p2)2=(2T+2mp)2≥(2mp+mK)2
We conclude that
(3.523) 2 T ≥ m K (3.523) 2 T ≥ m K {:(3.523)2T >= m_(K):}\begin{equation*} 2 T \geq m_{K} \tag{3.523} \end{equation*}(3.523)2T≥mK
and, thus, the total threshold kinetic energy is given by
(3.524) T th , tot = m K (3.524) T th , tot = m K {:(3.524)T_(th,tot)=m_(K):}\begin{equation*} T_{\mathrm{th}, \mathrm{tot}}=m_{K} \tag{3.524} \end{equation*}(3.524)Tth,tot=mK

1.109

By conservation of 4-momentum, we have the relation
(3.525) p χ + p p = k p + k χ ∗ (3.525) p χ + p p = k p + k χ ∗ {:(3.525)p_(chi)+p_(p)=k_(p)+k_(chi^(**)):}\begin{equation*} p_{\chi}+p_{p}=k_{p}+k_{\chi^{*}} \tag{3.525} \end{equation*}(3.525)pχ+pp=kp+kχ∗
where the p i p i p_(i)p_{i}pi are the incoming and the k i k i k_(i)k_{i}ki the outgoing 4-momenta. By squaring this relation, we find that
p χ 2 + 2 p χ ⋅ p p + p p 2 = m χ 2 + m p 2 + 2 ( m χ + T ) m p (3.526) = ( k p + k χ ) 2 ≥ ( m p + m χ + δ ) 2 p χ 2 + 2 p χ ⋅ p p + p p 2 = m χ 2 + m p 2 + 2 m χ + T m p (3.526) = k p + k χ 2 ≥ m p + m χ + δ 2 {:[p_(chi)^(2)+2p_(chi)*p_(p)+p_(p)^(2)=m_(chi)^(2)+m_(p)^(2)+2(m_(chi)+T)m_(p)],[(3.526)=(k_(p)+k_(chi))^(2) >= (m_(p)+m_(chi)+delta)^(2)]:}\begin{align*} p_{\chi}^{2}+2 p_{\chi} \cdot p_{p}+p_{p}^{2} & =m_{\chi}^{2}+m_{p}^{2}+2\left(m_{\chi}+T\right) m_{p} \\ & =\left(k_{p}+k_{\chi}\right)^{2} \geq\left(m_{p}+m_{\chi}+\delta\right)^{2} \tag{3.526} \end{align*}pχ2+2pχ⋅pp+pp2=mχ2+mp2+2(mχ+T)mp(3.526)=(kp+kχ)2≥(mp+mχ+δ)2
Rearranging this inequality leads to
(3.527) T ≥ ( 1 + m χ m p ) δ + δ 2 2 m p (3.527) T ≥ 1 + m χ m p δ + δ 2 2 m p {:(3.527)T >= (1+(m_(chi))/(m_(p)))delta+(delta^(2))/(2m_(p)):}\begin{equation*} T \geq\left(1+\frac{m_{\chi}}{m_{p}}\right) \delta+\frac{\delta^{2}}{2 m_{p}} \tag{3.527} \end{equation*}(3.527)T≥(1+mχmp)δ+δ22mp
In the limit δ ≪ m χ δ ≪ m χ delta≪m_(chi)\delta \ll m_{\chi}δ≪mχ, we can neglect the δ 2 δ 2 delta^(2)\delta^{2}δ2 term and obtain
(3.528) T ≥ ( 1 + m χ m p ) δ (3.528) T ≥ 1 + m χ m p δ {:(3.528)T >= (1+(m_(chi))/(m_(p)))delta:}\begin{equation*} T \geq\left(1+\frac{m_{\chi}}{m_{p}}\right) \delta \tag{3.528} \end{equation*}(3.528)T≥(1+mχmp)δ
As long as δ / T ≪ m χ / m p δ / T ≪ m χ / m p delta//T≪m_(chi)//m_(p)\delta / T \ll m_{\chi} / m_{p}δ/T≪mχ/mp, the threshold energy T T TTT will be in the classical limit and given by T = m χ v 2 / 2 T = m χ v 2 / 2 T=m_(chi)v^(2)//2T=m_{\chi} v^{2} / 2T=mχv2/2. This leads to
(3.529) v ≥ 2 m p + m χ m p m χ δ = 2 δ μ (3.529) v ≥ 2 m p + m χ m p m χ δ = 2 δ μ {:(3.529)v >= sqrt(2(m_(p)+m_(chi))/(m_(p)m_(chi)))delta=sqrt((2delta)/(mu)):}\begin{equation*} v \geq \sqrt{2 \frac{m_{p}+m_{\chi}}{m_{p} m_{\chi}}} \delta=\sqrt{\frac{2 \delta}{\mu}} \tag{3.529} \end{equation*}(3.529)v≥2mp+mχmpmχδ=2δμ
where μ ≡ m p m χ / ( m p + m χ ) μ ≡ m p m χ / m p + m χ mu-=m_(p)m_(chi)//(m_(p)+m_(chi))\mu \equiv m_{p} m_{\chi} /\left(m_{p}+m_{\chi}\right)μ≡mpmχ/(mp+mχ) is the reduced mass of the proton- χ χ chi\chiχ system.

1.110

Since, for a particle of mass m , p μ = ( E , p ) μ = m V μ = m γ ( 1 , v ) μ m , p μ = ( E , p ) μ = m V μ = m γ ( 1 , v ) μ m,p^(mu)=(E,p)^(mu)=mV^(mu)=m gamma(1,v)^(mu)m, p^{\mu}=(E, \boldsymbol{p})^{\mu}=m V^{\mu}=m \gamma(1, \boldsymbol{v})^{\mu}m,pμ=(E,p)μ=mVμ=mγ(1,v)μ, it follows that
(3.530) v = p E , v = | p | E (3.530) v = p E , v = | p | E {:(3.530)v=(p)/(E)","quad v=(|p|)/(E):}\begin{equation*} v=\frac{p}{E}, \quad v=\frac{|p|}{E} \tag{3.530} \end{equation*}(3.530)v=pE,v=|p|E
It also holds that
(3.531) m 2 = p 2 = E 2 − p 2 ⟹ p = E 1 − m 2 E 2 (3.531) m 2 = p 2 = E 2 − p 2 ⟹ p = E 1 − m 2 E 2 {:(3.531)m^(2)=p^(2)=E^(2)-p^(2)quad Longrightarrowquad p=Esqrt(1-(m^(2))/(E^(2))):}\begin{equation*} m^{2}=p^{2}=E^{2}-p^{2} \quad \Longrightarrow \quad p=E \sqrt{1-\frac{m^{2}}{E^{2}}} \tag{3.531} \end{equation*}(3.531)m2=p2=E2−p2⟹p=E1−m2E2
Thus, the time t ( m ) t ( m ) t(m)t(m)t(m) for neutrinos of mass m m mmm and energy E E EEE to reach the Earth is given by
(3.532) t ( m ) = L v = L 1 − m 2 E 2 (3.532) t ( m ) = L v = L 1 − m 2 E 2 {:(3.532)t(m)=(L)/(v)=(L)/(sqrt(1-(m^(2))/(E^(2)))):}\begin{equation*} t(m)=\frac{L}{v}=\frac{L}{\sqrt{1-\frac{m^{2}}{E^{2}}}} \tag{3.532} \end{equation*}(3.532)t(m)=Lv=L1−m2E2
and the difference compared to if they were massless by
(3.533) Δ t ≡ t ( m ) − t ( 0 ) = L ( 1 1 − m 2 E 2 − 1 ) ≃ m 2 L 2 E 2 (3.533) Δ t ≡ t ( m ) − t ( 0 ) = L 1 1 − m 2 E 2 − 1 ≃ m 2 L 2 E 2 {:(3.533)Delta t-=t(m)-t(0)=L((1)/(sqrt(1-(m^(2))/(E^(2))))-1)≃(m^(2)L)/(2E^(2)):}\begin{equation*} \Delta t \equiv t(m)-t(0)=L\left(\frac{1}{\sqrt{1-\frac{m^{2}}{E^{2}}}}-1\right) \simeq \frac{m^{2} L}{2 E^{2}} \tag{3.533} \end{equation*}(3.533)Δt≡t(m)−t(0)=L(11−m2E2−1)≃m2L2E2
where the last approximation holds when m ≪ E m ≪ E m≪Em \ll Em≪E.

1.111

a) From conservation of energy, it follows that the total energy of the elementary particle after acceleration is given by
(3.534) E = E in + E acc (3.534) E = E in + E acc {:(3.534)E=E_(in)+E_(acc):}\begin{equation*} E=E_{\mathrm{in}}+E_{\mathrm{acc}} \tag{3.534} \end{equation*}(3.534)E=Ein+Eacc
where E in E in  E_("in ")E_{\text {in }}Ein  is the initial total energy, and E acc E acc  E_("acc ")E_{\text {acc }}Eacc  is the energy added by acceleration. Since the particle was initially at rest, we have E in = m c 2 E in  = m c 2 E_("in ")=mc^(2)E_{\text {in }}=m c^{2}Ein =mc2, where m m mmm is the mass of the particle. From the relation
(3.535) E = m c 2 + T (3.535) E = m c 2 + T {:(3.535)E=mc^(2)+T:}\begin{equation*} E=m c^{2}+T \tag{3.535} \end{equation*}(3.535)E=mc2+T
where T T TTT is the kinetic energy, it immediately follows that T = E acc. T = E acc.  T=E_("acc. ")T=E_{\text {acc. }}T=Eacc. . The energy added by the accelerator is given by
(3.536) E acc = Q Δ U (3.536) E acc = Q Δ U {:(3.536)E_(acc)=Q Delta U:}\begin{equation*} E_{\mathrm{acc}}=Q \Delta U \tag{3.536} \end{equation*}(3.536)Eacc=QΔU
where Q = e Q = e Q=eQ=eQ=e is the charge of the particle and Δ U Δ U Delta U\Delta UΔU is the difference in the electric potential at the start and end of the acceleration. Since the electric field is known, we can easily compute Δ U Δ U Delta U\Delta UΔU as
(3.537) Δ U = E ′ L = 10 4 V / m ⋅ 100 m = 10 6 V (3.537) Δ U = E ′ L = 10 4 V / m ⋅ 100 m = 10 6 V {:(3.537)Delta U=E^(')L=10^(4)V//m*100m=10^(6)V:}\begin{equation*} \Delta U=E^{\prime} L=10^{4} \mathrm{~V} / \mathrm{m} \cdot 100 \mathrm{~m}=10^{6} \mathrm{~V} \tag{3.537} \end{equation*}(3.537)ΔU=E′L=104 V/m⋅100 m=106 V
where E ′ E ′ E^(')E^{\prime}E′ is the electric field strength. Thus, the kinetic energy after the acceleration is
(3.538) T = e Δ U = 10 6 eV = 1 MeV (3.538) T = e Δ U = 10 6 eV = 1 MeV {:(3.538)T=e Delta U=10^(6)eV=1MeV:}\begin{equation*} T=e \Delta U=10^{6} \mathrm{eV}=1 \mathrm{MeV} \tag{3.538} \end{equation*}(3.538)T=eΔU=106eV=1MeV
b) The time for the particle to travel the distance d x d x dxd xdx in the tube is given by
(3.539) d t = d t d x d x = 1 v d x (3.539) d t = d t d x d x = 1 v d x {:(3.539)dt=(dt)/(dx)dx=(1)/(v)dx:}\begin{equation*} d t=\frac{d t}{d x} d x=\frac{1}{v} d x \tag{3.539} \end{equation*}(3.539)dt=dtdxdx=1vdx
Thus, the total time for the particle to pass through the tube is
(3.540) t = ∫ 0 L 1 v d x (3.540) t = ∫ 0 L   1 v d x {:(3.540)t=int_(0)^(L)(1)/(v)dx:}\begin{equation*} t=\int_{0}^{L} \frac{1}{v} d x \tag{3.540} \end{equation*}(3.540)t=∫0L1vdx
where L L LLL is the detector length. However, the length traveled in the detector is related to the total energy as
(3.541) E = e E ′ x + m c 2 (3.541) E = e E ′ x + m c 2 {:(3.541)E=eE^(')x+mc^(2):}\begin{equation*} E=e E^{\prime} x+m c^{2} \tag{3.541} \end{equation*}(3.541)E=eE′x+mc2
cf., problem a). Thus, we change the variable of integration to the total energy E E EEE and obtain
(3.542) t = ∫ m c 2 m c 2 + E acc d E v e E ′ (3.542) t = ∫ m c 2 m c 2 + E acc   d E v e E ′ {:(3.542)t=int_(mc^(2))^(mc^(2)+E_(acc))(dE)/(veE^(')):}\begin{equation*} t=\int_{m c^{2}}^{m c^{2}+E_{\mathrm{acc}}} \frac{d E}{v e E^{\prime}} \tag{3.542} \end{equation*}(3.542)t=∫mc2mc2+EaccdEveE′
The velocity v v vvv can be expressed as a function of the total energy through the relation
(3.543) v = p c 2 E = c 1 − m 2 c 4 E 2 (3.543) v = p c 2 E = c 1 − m 2 c 4 E 2 {:(3.543)v=(pc^(2))/(E)=csqrt(1-(m^(2)c^(4))/(E^(2))):}\begin{equation*} v=\frac{p c^{2}}{E}=c \sqrt{1-\frac{m^{2} c^{4}}{E^{2}}} \tag{3.543} \end{equation*}(3.543)v=pc2E=c1−m2c4E2
which yields
(3.544) t = 1 c e E ′ ∫ m c 2 m c 2 + E acc E d E E 2 − m 2 c 4 (3.544) t = 1 c e E ′ ∫ m c 2 m c 2 + E acc   E d E E 2 − m 2 c 4 {:(3.544)t=(1)/(ceE^('))int_(mc^(2))^(mc^(2)+E_(acc))(EdE)/(sqrt(E^(2)-m^(2)c^(4))):}\begin{equation*} t=\frac{1}{c e E^{\prime}} \int_{m c^{2}}^{m c^{2}+E_{\mathrm{acc}}} \frac{E d E}{\sqrt{E^{2}-m^{2} c^{4}}} \tag{3.544} \end{equation*}(3.544)t=1ceE′∫mc2mc2+EaccEdEE2−m2c4
when inserted into the integral. A new change of variables to α = E 2 α = E 2 alpha=E^(2)\alpha=E^{2}α=E2 results in the formula
t = 1 c e E ′ ∫ m 2 c 4 ( m c 2 + E acc ) 2 d α 2 α − m 2 c 4 = 1 c e E ′ [ α − m 2 c 4 ] m 2 c 4 ( m c 2 + E acc ) 2 (3.545) = 1 c e E ′ E acc ( 2 m c 2 + E acc ) = 1 c e E ′ , e E ′ L ( 2 m c 2 + e E ′ L ) t = 1 c e E ′ ∫ m 2 c 4 m c 2 + E acc 2   d α 2 α − m 2 c 4 = 1 c e E ′ α − m 2 c 4 m 2 c 4 m c 2 + E acc 2 (3.545) = 1 c e E ′ E acc 2 m c 2 + E acc = 1 c e E ′ , e E ′ L 2 m c 2 + e E ′ L {:[t=(1)/(ceE^('))int_(m^(2)c^(4))^((mc^(2)+E_(acc))^(2))(d alpha)/(2sqrt(alpha-m^(2)c^(4)))=(1)/(ceE^('))[sqrt(alpha-m^(2)c^(4))]_(m^(2)c^(4))^((mc^(2)+E_(acc))^(2))],[(3.545){:=(1)/(ceE^('))sqrt(E_(acc)(2mc^(2)+E_(acc)))=(1)/(ceE^(')),sqrt(eE^(')L(2mc^(2)+eE^(')L:}))]:}\begin{align*} t & =\frac{1}{c e E^{\prime}} \int_{m^{2} c^{4}}^{\left(m c^{2}+E_{\mathrm{acc}}\right)^{2}} \frac{d \alpha}{2 \sqrt{\alpha-m^{2} c^{4}}}=\frac{1}{c e E^{\prime}}\left[\sqrt{\alpha-m^{2} c^{4}}\right]_{m^{2} c^{4}}^{\left(m c^{2}+E_{\mathrm{acc}}\right)^{2}} \\ & \left.=\frac{1}{c e E^{\prime}} \sqrt{E_{\mathrm{acc}}\left(2 m c^{2}+E_{\mathrm{acc}}\right)}=\frac{1}{c e E^{\prime}}, \sqrt{e E^{\prime} L\left(2 m c^{2}+e E^{\prime} L\right.}\right) \tag{3.545} \end{align*}t=1ceE′∫m2c4(mc2+Eacc)2dα2α−m2c4=1ceE′[α−m2c4]m2c4(mc2+Eacc)2(3.545)=1ceE′Eacc(2mc2+Eacc)=1ceE′,eE′L(2mc2+eE′L)

1.112

Let A μ = A μ ( x ) A μ = A μ ( x ) A^(mu)=A^(mu)(x)A^{\mu}=A^{\mu}(x)Aμ=Aμ(x) be the solution to ◻ A μ ( x ) = 0 ◻ A μ ( x ) = 0 ◻A^(mu)(x)=0\square A^{\mu}(x)=0◻Aμ(x)=0. Calculate ◻ ′ A ′ μ ( x ′ ) ◻ ′ A ′ μ x ′ ◻^(')A^('mu)(x^('))\square^{\prime} A^{\prime \mu}\left(x^{\prime}\right)◻′A′μ(x′), where x ′ = Λ x x ′ = Λ x x^(')=Lambda xx^{\prime}=\Lambda xx′=Λx and A ′ μ ( x ′ ) = Λ μ v A v ( x ) A ′ μ x ′ = Λ μ v A v ( x ) A^('mu)(x^('))=Lambda^(mu)_(v)A^(v)(x)A^{\prime \mu}\left(x^{\prime}\right)=\Lambda^{\mu}{ }_{v} A^{v}(x)A′μ(x′)=ΛμvAv(x). An explicit calculation yields
A ′ μ ( x ′ ) = ∂ v ′ ∂ ′ v A ′ μ = Λ v α ∂ α Λ β v ∂ β Λ γ μ A γ ( x ) = Λ v α Λ β v Λ γ μ ∂ α ∂ β A γ = ( Λ T ) α v Λ v β Λ μ γ ∂ α ∂ β A γ = ( Λ T Λ ) α β Λ γ μ ∂ α ∂ β A γ = η β α Λ γ μ ∂ α ∂ β A γ (3.546) = Λ γ μ γ ∂ α ∂ α A γ = Λ γ μ ◻ A γ = { ◻ A μ = 0 } = Λ γ μ ⋅ 0 = 0 , A ′ μ x ′ = ∂ v ′ ∂ ′ v A ′ μ = Λ v α ∂ α Λ β v ∂ β Λ γ μ A γ ( x ) = Λ v α Λ β v Λ γ μ ∂ α ∂ β A γ = Λ T α v Λ v β Λ μ γ ∂ α ∂ β A γ = Λ T Λ α β Λ γ μ ∂ α ∂ β A γ = η β α Λ γ μ ∂ α ∂ β A γ (3.546) = Λ γ μ γ ∂ α ∂ α A γ = Λ γ μ ◻ A γ = ◻ A μ = 0 = Λ γ μ ⋅ 0 = 0 , {:[A^('mu)(x^('))=del_(v)^(')del^('v)A^('mu)=Lambda_(v)^(alpha)del_(alpha)Lambda_(beta)^(v)del^(beta)Lambda_(gamma)^(mu)A^(gamma)(x)=Lambda_(v)^(alpha)Lambda_(beta)^(v)Lambda_(gamma)^(mu)del_(alpha)del^(beta)A^(gamma)],[=(Lambda^(T))^(alpha)_(v)Lambda^(v)_(beta)Lambda^(mu)_(gamma)del_(alpha)del^(beta)A^(gamma)=(Lambda^(T)Lambda)^(alpha)_(beta)Lambda_(gamma)^(mu)del_(alpha)del^(beta)A^(gamma)=eta_(beta)^(alpha)Lambda_(gamma)^(mu)del_(alpha)del^(beta)A^(gamma)],[(3.546)=Lambda_(gamma)^(mu)_(gamma)del_(alpha)del^(alpha)A^(gamma)=Lambda_(gamma)^(mu)◻A^(gamma)={◻A^(mu)=0}=Lambda_(gamma)^(mu)*0=0","]:}\begin{align*} A^{\prime \mu}\left(x^{\prime}\right) & =\partial_{v}^{\prime} \partial^{\prime v} A^{\prime \mu}=\Lambda_{v}^{\alpha} \partial_{\alpha} \Lambda_{\beta}^{v} \partial^{\beta} \Lambda_{\gamma}^{\mu} A^{\gamma}(x)=\Lambda_{v}^{\alpha} \Lambda_{\beta}^{v} \Lambda_{\gamma}^{\mu} \partial_{\alpha} \partial^{\beta} A^{\gamma} \\ & =\left(\Lambda^{T}\right)^{\alpha}{ }_{v} \Lambda^{v}{ }_{\beta} \Lambda^{\mu}{ }_{\gamma} \partial_{\alpha} \partial^{\beta} A^{\gamma}=\left(\Lambda^{T} \Lambda\right)^{\alpha}{ }_{\beta} \Lambda_{\gamma}^{\mu} \partial_{\alpha} \partial^{\beta} A^{\gamma}=\eta_{\beta}^{\alpha} \Lambda_{\gamma}^{\mu} \partial_{\alpha} \partial^{\beta} A^{\gamma} \\ & =\Lambda_{\gamma}^{\mu}{ }_{\gamma} \partial_{\alpha} \partial^{\alpha} A^{\gamma}=\Lambda_{\gamma}^{\mu} \square A^{\gamma}=\left\{\square A^{\mu}=0\right\}=\Lambda_{\gamma}^{\mu} \cdot 0=0, \tag{3.546} \end{align*}A′μ(x′)=∂v′∂′vA′μ=Λvα∂αΛβv∂βΛγμAγ(x)=ΛvαΛβvΛγμ∂α∂βAγ=(ΛT)αvΛvβΛμγ∂α∂βAγ=(ΛTΛ)αβΛγμ∂α∂βAγ=ηβαΛγμ∂α∂βAγ(3.546)=Λγμγ∂α∂αAγ=Λγμ◻Aγ={◻Aμ=0}=Λγμ⋅0=0,
i.e., transformations.

1.113

Inserting the definition A μ ′ = A μ + ∂ μ ψ A μ ′ = A μ + ∂ μ ψ A_(mu)^(')=A_(mu)+del_(mu)psiA_{\mu}^{\prime}=A_{\mu}+\partial_{\mu} \psiAμ′=Aμ+∂μψ into F μ ν ′ = ∂ μ A ν ′ − ∂ ν A μ ′ F μ ν ′ = ∂ μ A ν ′ − ∂ ν A μ ′ F_(mu nu)^(')=del_(mu)A_(nu)^(')-del_(nu)A_(mu)^(')F_{\mu \nu}^{\prime}=\partial_{\mu} A_{\nu}^{\prime}-\partial_{\nu} A_{\mu}^{\prime}Fμν′=∂μAν′−∂νAμ′, we obtain
(3.547) F μ ν ′ = ∂ μ A ν + ∂ μ ∂ ν ψ − ∂ v A μ − ∂ ν ∂ μ ψ = F μ ν + ( ∂ μ ∂ v − ∂ ν ∂ μ ) ψ ⏟ = 0 = F μ ν . (3.547) F μ ν ′ = ∂ μ A ν + ∂ μ ∂ ν ψ − ∂ v A μ − ∂ ν ∂ μ ψ = F μ ν + ∂ μ ∂ v − ∂ ν ∂ μ ψ ⏟ = 0 = F μ ν . {:(3.547)F_(mu nu)^(')=del_(mu)A_(nu)+del_(mu)del_(nu)psi-del_(v)A_(mu)-del_(nu)del_(mu)psi=F_(mu nu)+ubrace((del_(mu)del_(v)-del_(nu)del_(mu))psiubrace)_(=0)=F_(mu nu).:}\begin{equation*} F_{\mu \nu}^{\prime}=\partial_{\mu} A_{\nu}+\partial_{\mu} \partial_{\nu} \psi-\partial_{v} A_{\mu}-\partial_{\nu} \partial_{\mu} \psi=F_{\mu \nu}+\underbrace{\left(\partial_{\mu} \partial_{v}-\partial_{\nu} \partial_{\mu}\right) \psi}_{=0}=F_{\mu \nu} . \tag{3.547} \end{equation*}(3.547)Fμν′=∂μAν+∂μ∂νψ−∂vAμ−∂ν∂μψ=Fμν+(∂μ∂v−∂ν∂μ)ψ⏟=0=Fμν.
Thus, the gauge transformation results in the same field tensor F μ ν F μ ν F_(mu nu)F_{\mu \nu}Fμν.

1.114

The standard configuration Lorentz transformation is given by
(3.548) x ′ 0 = x 0 cosh ⁡ θ − x 1 sinh ⁡ θ (3.549) x ′ 1 = − x 0 sinh ⁡ θ + x 1 cosh ⁡ θ (3.548) x ′ 0 = x 0 cosh ⁡ θ − x 1 sinh ⁡ θ (3.549) x ′ 1 = − x 0 sinh ⁡ θ + x 1 cosh ⁡ θ {:[(3.548)x^('0)=x^(0)cosh theta-x^(1)sinh theta],[(3.549)x^('1)=-x^(0)sinh theta+x^(1)cosh theta]:}\begin{align*} & x^{\prime 0}=x^{0} \cosh \theta-x^{1} \sinh \theta \tag{3.548}\\ & x^{\prime 1}=-x^{0} \sinh \theta+x^{1} \cosh \theta \tag{3.549} \end{align*}(3.548)x′0=x0cosh⁡θ−x1sinh⁡θ(3.549)x′1=−x0sinh⁡θ+x1cosh⁡θ
(3.550) x ′ 2 = x 2 , (3.551) x ′ 3 = x 3 (3.550) x ′ 2 = x 2 , (3.551) x ′ 3 = x 3 {:[(3.550)x^('2)=x^(2)","],[(3.551)x^('3)=x^(3)]:}\begin{align*} & x^{\prime 2}=x^{2}, \tag{3.550}\\ & x^{\prime 3}=x^{3} \tag{3.551} \end{align*}(3.550)x′2=x2,(3.551)x′3=x3
where tanh ⁡ θ = v / c tanh ⁡ θ = v / c tanh theta=v//c\tanh \theta=v / ctanh⁡θ=v/c. This means that the Lorentz transformation in matrix from is
(3.552) Λ = ( cosh ⁡ θ − sinh ⁡ θ 0 0 − sinh ⁡ θ cosh ⁡ θ 0 0 0 0 1 0 0 0 0 1 ) (3.552) Λ = cosh ⁡ θ − sinh ⁡ θ 0 0 − sinh ⁡ θ cosh ⁡ θ 0 0 0 0 1 0 0 0 0 1 {:(3.552)Lambda=([cosh theta,-sinh theta,0,0],[-sinh theta,cosh theta,0,0],[0,0,1,0],[0,0,0,1]):}\Lambda=\left(\begin{array}{cccc} \cosh \theta & -\sinh \theta & 0 & 0 \tag{3.552}\\ -\sinh \theta & \cosh \theta & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)(3.552)Λ=(cosh⁡θ−sinh⁡θ00−sinh⁡θcosh⁡θ0000100001)
such that x ′ = Λ x x ′ = Λ x x^(')=Lambda xx^{\prime}=\Lambda xx′=Λx.
a) The observer in K ′ K ′ K^(')K^{\prime}K′ must measure the stick simultaneously in both endpoints. This means at time x ′ 0 = 0 x ′ 0 = 0 x^('0)=0x^{\prime 0}=0x′0=0 in his/her coordinate system. Without loss of generality, we can put one of the endpoints of the stick at the origin in K ′ K ′ K^(')K^{\prime}K′. Thus, we have ( 0 , 0 , 0 ) ( 0 , 0 , 0 ) (0,0,0)(0,0,0)(0,0,0) in K ′ K ′ K^(')K^{\prime}K′ at time x ′ 0 = 0 x ′ 0 = 0 x^('0)=0x^{\prime 0}=0x′0=0 and ( ℓ , 0 , 0 ) ( ℓ , 0 , 0 ) (ℓ,0,0)(\ell, 0,0)(ℓ,0,0) in K ′ K ′ K^(')K^{\prime}K′ at time x ′ 0 = 0 x ′ 0 = 0 x^('0)=0x^{\prime 0}=0x′0=0. Therefore, it holds that Δ x ′ 0 = 0 = Δ x 0 cosh ⁡ θ − ℓ sinh ⁡ θ Δ x ′ 0 = 0 = Δ x 0 cosh ⁡ θ − ℓ sinh ⁡ θ Deltax^('0)=0=Deltax^(0)cosh theta-ℓsinh theta\Delta x^{\prime 0}=0=\Delta x^{0} \cosh \theta-\ell \sinh \thetaΔx′0=0=Δx0cosh⁡θ−ℓsinh⁡θ, which leads to
(3.553) Δ x 0 = ℓ sinh ⁡ θ cosh ⁡ θ = ℓ tanh ⁡ θ (3.553) Δ x 0 = ℓ sinh ⁡ θ cosh ⁡ θ = ℓ tanh ⁡ θ {:(3.553)Deltax^(0)=(ℓsinh theta)/(cosh theta)=ℓtanh theta:}\begin{equation*} \Delta x^{0}=\frac{\ell \sinh \theta}{\cosh \theta}=\ell \tanh \theta \tag{3.553} \end{equation*}(3.553)Δx0=ℓsinh⁡θcosh⁡θ=ℓtanh⁡θ
However, ℓ ′ = − Δ x 0 sinh ⁡ θ + ℓ cosh ⁡ θ ℓ ′ = − Δ x 0 sinh ⁡ θ + ℓ cosh ⁡ θ ℓ^(')=-Deltax^(0)sinh theta+ℓcosh theta\ell^{\prime}=-\Delta x^{0} \sinh \theta+\ell \cosh \thetaℓ′=−Δx0sinh⁡θ+ℓcosh⁡θ, so we have
ℓ ′ = − ℓ sinh ⁡ θ cosh ⁡ θ sinh ⁡ θ + ℓ cosh ⁡ θ = ℓ cosh 2 ⁡ θ − sinh 2 ⁡ θ cosh ⁡ θ = { cosh 2 ⁡ θ − sinh 2 ⁡ θ = 1 } (3.554) = ℓ cosh ⁡ θ ℓ ′ = − ℓ sinh ⁡ θ cosh ⁡ θ sinh ⁡ θ + ℓ cosh ⁡ θ = ℓ cosh 2 ⁡ θ − sinh 2 ⁡ θ cosh ⁡ θ = cosh 2 ⁡ θ − sinh 2 ⁡ θ = 1 (3.554) = ℓ cosh ⁡ θ {:[ℓ^(')=-(ℓsinh theta)/(cosh theta)sinh theta+ℓcosh theta=ℓ(cosh^(2)theta-sinh^(2)theta)/(cosh theta)={cosh^(2)theta-sinh^(2)theta=1}],[(3.554)=(ℓ)/(cosh theta)]:}\begin{align*} \ell^{\prime} & =-\frac{\ell \sinh \theta}{\cosh \theta} \sinh \theta+\ell \cosh \theta=\ell \frac{\cosh ^{2} \theta-\sinh ^{2} \theta}{\cosh \theta}=\left\{\cosh ^{2} \theta-\sinh ^{2} \theta=1\right\} \\ & =\frac{\ell}{\cosh \theta} \tag{3.554} \end{align*}ℓ′=−ℓsinh⁡θcosh⁡θsinh⁡θ+ℓcosh⁡θ=ℓcosh2⁡θ−sinh2⁡θcosh⁡θ={cosh2⁡θ−sinh2⁡θ=1}(3.554)=ℓcosh⁡θ
The relation cosh ⁡ θ = 1 / 1 − v 2 / c 2 cosh ⁡ θ = 1 / 1 − v 2 / c 2 cosh theta=1//sqrt(1-v^(2)//c^(2))\cosh \theta=1 / \sqrt{1-v^{2} / c^{2}}cosh⁡θ=1/1−v2/c2 implies that ℓ ′ = ℓ 1 − v 2 / c 2 ℓ ′ = ℓ 1 − v 2 / c 2 ℓ^(')=ℓsqrt(1-v^(2)//c^(2))\ell^{\prime}=\ell \sqrt{1-v^{2} / c^{2}}ℓ′=ℓ1−v2/c2. Thus, the result is
(3.555) Δ x ′ = ( ℓ ′ , 0 , 0 ) = ( ℓ 1 − v 2 / c 2 , 0 , 0 ) (3.555) Δ x ′ = ℓ ′ , 0 , 0 = ℓ 1 − v 2 / c 2 , 0 , 0 {:(3.555)Deltax^(')=(ℓ^('),0,0)=(ℓsqrt(1-v^(2)//c^(2)),0,0):}\begin{equation*} \Delta \mathbf{x}^{\prime}=\left(\ell^{\prime}, 0,0\right)=\left(\ell \sqrt{1-v^{2} / c^{2}}, 0,0\right) \tag{3.555} \end{equation*}(3.555)Δx′=(ℓ′,0,0)=(ℓ1−v2/c2,0,0)
b) The electric and magnetic fields E = ( 0 , 0 , E ) E = ( 0 , 0 , E ) E=(0,0,E)\mathbf{E}=(0,0, E)E=(0,0,E) and B = ( 0 , 0 , 0 ) B = ( 0 , 0 , 0 ) B=(0,0,0)\mathbf{B}=(0,0,0)B=(0,0,0) in K K KKK implies that the electromagnetic field strength tensor in K K KKK is
(3.556) F = ( 0 0 0 − E 0 0 0 0 0 0 0 0 E 0 0 0 ) (3.556) F = 0 0 0 − E 0 0 0 0 0 0 0 0 E 0 0 0 {:(3.556)F=([0,0,0,-E],[0,0,0,0],[0,0,0,0],[E,0,0,0]):}F=\left(\begin{array}{cccc} 0 & 0 & 0 & -E \tag{3.556}\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ E & 0 & 0 & 0 \end{array}\right)(3.556)F=(000−E00000000E000)
The electromagnetic field strength tensor in K ′ K ′ K^(')K^{\prime}K′ is given by F ′ = Λ F Λ T F ′ = Λ F Λ T F^(')=Lambda FLambda^(T)F^{\prime}=\Lambda F \Lambda^{T}F′=ΛFΛT. Thus, we find
F ′ = Λ F Λ T = ( cosh ⁡ θ − sinh ⁡ θ 0 0 − sinh ⁡ θ cosh ⁡ θ 0 0 0 0 1 0 0 0 0 1 ) ( 0 0 0 − E 0 0 0 0 0 0 0 0 E 0 0 0 ) × ( cosh ⁡ θ − sinh ⁡ θ 0 0 − sinh ⁡ θ cosh ⁡ θ 0 0 0 0 1 0 0 0 0 1 ) = ( 0 0 0 − E cosh ⁡ θ 0 0 0 E sinh ⁡ θ 0 0 0 0 E cosh ⁡ θ − E sinh ⁡ θ 0 0 ) (3.557) = ( 0 − E ′ 1 − E ′ 2 − E ′ 3 E ′ 1 0 − c B ′ 3 c B ′ 2 E ′ 2 c B ′ 3 0 − c B ′ 2 E ′ 3 − c B ′ 2 c B ′ 1 0 . ) F ′ = Λ F Λ T = cosh ⁡ θ − sinh ⁡ θ 0 0 − sinh ⁡ θ cosh ⁡ θ 0 0 0 0 1 0 0 0 0 1 0 0 0 − E 0 0 0 0 0 0 0 0 E 0 0 0 × cosh ⁡ θ − sinh ⁡ θ 0 0 − sinh ⁡ θ cosh ⁡ θ 0 0 0 0 1 0 0 0 0 1 = 0 0 0 − E cosh ⁡ θ 0 0 0 E sinh ⁡ θ 0 0 0 0 E cosh ⁡ θ − E sinh ⁡ θ 0 0 (3.557) = 0 − E ′ 1 − E ′ 2 − E ′ 3 E ′ 1 0 − c B ′ 3 c B ′ 2 E ′ 2 c B ′ 3 0 − c B ′ 2 E ′ 3 − c B ′ 2 c B ′ 1 0 . {:[F^(')=Lambda FLambda^(T)=([cosh theta,-sinh theta,0,0],[-sinh theta,cosh theta,0,0],[0,0,1,0],[0,0,0,1])([0,0,0,-E],[0,0,0,0],[0,0,0,0],[E,0,0,0])],[ xx([cosh theta,-sinh theta,0,0],[-sinh theta,cosh theta,0,0],[0,0,1,0],[0,0,0,1])],[=([0,0,0,-E cosh theta],[0,0,0,E sinh theta],[0,0,0,0],[E cosh theta,-E sinh theta,0,0])],[(3.557)=([0,-E^('1),-E^('2),-E^('3)],[E^('1),0,-cB^('3),cB^('2)],[E^('2),cB^('3),0,-cB^('2)],[E^('3),-cB^('2),cB^('1),0.])]:}\begin{align*} F^{\prime} & =\Lambda F \Lambda^{T}=\left(\begin{array}{cccc} \cosh \theta & -\sinh \theta & 0 & 0 \\ -\sinh \theta & \cosh \theta & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)\left(\begin{array}{llll} 0 & 0 & 0 & -E \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ E & 0 & 0 & 0 \end{array}\right) \\ & \times\left(\begin{array}{cccc} \cosh \theta & -\sinh \theta & 0 & 0 \\ -\sinh \theta & \cosh \theta & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right) \\ & =\left(\begin{array}{cccc} 0 & 0 & 0 & -E \cosh \theta \\ 0 & 0 & 0 & E \sinh \theta \\ 0 & 0 & 0 & 0 \\ E \cosh \theta & -E \sinh \theta & 0 & 0 \end{array}\right) \\ & =\left(\begin{array}{cccc} 0 & -E^{\prime 1} & -E^{\prime 2} & -E^{\prime 3} \\ E^{\prime 1} & 0 & -c B^{\prime 3} & c B^{\prime 2} \\ E^{\prime 2} & c B^{\prime 3} & 0 & -c B^{\prime 2} \\ E^{\prime 3} & -c B^{\prime 2} & c B^{\prime 1} & 0 . \end{array}\right) \tag{3.557} \end{align*}F′=ΛFΛT=(cosh⁡θ−sinh⁡θ00−sinh⁡θcosh⁡θ0000100001)(000−E00000000E000)×(cosh⁡θ−sinh⁡θ00−sinh⁡θcosh⁡θ0000100001)=(000−Ecosh⁡θ000Esinh⁡θ0000Ecosh⁡θ−Esinh⁡θ00)(3.557)=(0−E′1−E′2−E′3E′10−cB′3cB′2E′2cB′30−cB′2E′3−cB′2cB′10.)
Therefore, E ′ 1 = E ′ 2 = 0 , E ′ 3 = E cosh ⁡ θ , B ′ 1 = B ′ 3 = 0 E ′ 1 = E ′ 2 = 0 , E ′ 3 = E cosh ⁡ θ , B ′ 1 = B ′ 3 = 0 E^('1)=E^('2)=0,E^('3)=E cosh theta,B^('1)=B^('3)=0E^{\prime 1}=E^{\prime 2}=0, E^{\prime 3}=E \cosh \theta, B^{\prime 1}=B^{\prime 3}=0E′1=E′2=0,E′3=Ecosh⁡θ,B′1=B′3=0, and B ′ 2 = E c sinh ⁡ θ B ′ 2 = E c sinh ⁡ θ B^('2)=(E)/(c)sinh thetaB^{\prime 2}=\frac{E}{c} \sinh \thetaB′2=Ecsinh⁡θ. Using the relations cosh ⁡ θ = γ ( v ) cosh ⁡ θ = γ ( v ) cosh theta=gamma(v)\cosh \theta=\gamma(v)cosh⁡θ=γ(v) and sinh ⁡ θ = v c γ ( v ) sinh ⁡ θ = v c γ ( v ) sinh theta=(v)/(c)gamma(v)\sinh \theta=\frac{v}{c} \gamma(v)sinh⁡θ=vcγ(v), where γ ( v ) = 1 1 − ( v c ) 2 γ ( v ) = 1 1 − v c 2 gamma(v)=(1)/(sqrt(1-((v)/(c))^(2)))\gamma(v)=\frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^{2}}}γ(v)=11−(vc)2, we obtain the electric and magnetic fields in K ′ K ′ K^(')K^{\prime}K′ as
(3.558) E ′ = ( E ′ 1 , E ′ 2 , E ′ 3 ) = ( 0 , 0 , E γ ( v ) ) (3.559) B ′ = ( B ′ 1 , B ′ 2 , B ′ 3 ) = ( 0 , E v c 2 γ ( v ) , 0 ) . (3.558) E ′ = E ′ 1 , E ′ 2 , E ′ 3 = ( 0 , 0 , E γ ( v ) ) (3.559) B ′ = B ′ 1 , B ′ 2 , B ′ 3 = 0 , E v c 2 γ ( v ) , 0 . {:[(3.558)E^(')=(E^('1),E^('2),E^('3))=(0","0","E gamma(v))],[(3.559)B^(')=(B^('1),B^('2),B^('3))=(0,E(v)/(c^(2))gamma(v),0).]:}\begin{align*} & \mathbf{E}^{\prime}=\left(E^{\prime 1}, E^{\prime 2}, E^{\prime 3}\right)=(0,0, E \gamma(v)) \tag{3.558}\\ & \mathbf{B}^{\prime}=\left(B^{\prime 1}, B^{\prime 2}, B^{\prime 3}\right)=\left(0, E \frac{v}{c^{2}} \gamma(v), 0\right) . \tag{3.559} \end{align*}(3.558)E′=(E′1,E′2,E′3)=(0,0,Eγ(v))(3.559)B′=(B′1,B′2,B′3)=(0,Evc2γ(v),0).

1.115

For small velocities, i.e., β = v / c ≪ 1 β = v / c ≪ 1 beta=v//c≪1\beta=v / c \ll 1β=v/c≪1, we have
(3.560) γ ≃ 1 + 1 2 v 2 c 2 ≃ 1 , β γ ≃ v c ( 1 + 1 2 ⋅ v 2 c 2 ) ≃ v c (3.560) γ ≃ 1 + 1 2 v 2 c 2 ≃ 1 , β γ ≃ v c 1 + 1 2 ⋅ v 2 c 2 ≃ v c {:(3.560)gamma≃1+(1)/(2)(v^(2))/(c^(2))≃1","quad beta gamma≃(v)/(c)(1+(1)/(2)*(v^(2))/(c^(2)))≃(v)/(c):}\begin{equation*} \gamma \simeq 1+\frac{1}{2} \frac{v^{2}}{c^{2}} \simeq 1, \quad \beta \gamma \simeq \frac{v}{c}\left(1+\frac{1}{2} \cdot \frac{v^{2}}{c^{2}}\right) \simeq \frac{v}{c} \tag{3.560} \end{equation*}(3.560)γ≃1+12v2c2≃1,βγ≃vc(1+12⋅v2c2)≃vc
Consider the electric and magnetic fields in K K KKK, i.e., E = ( E 1 , E 2 , E 3 ) E = E 1 , E 2 , E 3 E=(E^(1),E^(2),E^(3))\mathbf{E}=\left(E^{1}, E^{2}, E^{3}\right)E=(E1,E2,E3) and B = 0 B = 0 B=0\mathbf{B}=\mathbf{0}B=0. Thus, the corresponding electromagnetic field strength tensor in K ′ K ′ K^(')K^{\prime}K′ is given by F ′ = Λ F Λ T F ′ = Λ F Λ T F^(')=Lambda FLambda^(T)F^{\prime}=\Lambda F \Lambda^{T}F′=ΛFΛT, where
F = ( F μ ν ) = ( 0 − E 1 − E 2 − E 3 E 1 0 0 0 E 2 0 0 0 E 3 0 0 0 ) , Λ = ( Λ μ ν ) ≃ ( 1 − β 0 0 − β 1 0 0 0 0 1 0 0 0 0 1 ) F = F μ ν = 0 − E 1 − E 2 − E 3 E 1 0 0 0 E 2 0 0 0 E 3 0 0 0 , Λ = Λ μ ν ≃ 1 − β 0 0 − β 1 0 0 0 0 1 0 0 0 0 1 F=(F^(mu nu))=([0,-E^(1),-E^(2),-E^(3)],[E^(1),0,0,0],[E^(2),0,0,0],[E^(3),0,0,0]),quad Lambda=(Lambda^(mu)_(nu))≃([1,-beta,0,0],[-beta,1,0,0],[0,0,1,0],[0,0,0,1])F=\left(F^{\mu \nu}\right)=\left(\begin{array}{cccc}0 & -E^{1} & -E^{2} & -E^{3} \\ E^{1} & 0 & 0 & 0 \\ E^{2} & 0 & 0 & 0 \\ E^{3} & 0 & 0 & 0\end{array}\right), \quad \Lambda=\left(\Lambda^{\mu}{ }_{\nu}\right) \simeq\left(\begin{array}{cccc}1 & -\beta & 0 & 0 \\ -\beta & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right)F=(Fμν)=(0−E1−E2−E3E1000E2000E3000),Λ=(Λμν)≃(1−β00−β10000100001),
are the electromagnetic field strength tensor in K K KKK and the Lorentz transformation in the x 1 x 1 x^(1)x^{1}x1-direction with velocity v v vvv, respectively. Therefore, we find that
(3.562) F ′ ≃ ( 0 − E 1 − E 2 − E 3 E 1 0 β E 2 β E 3 E 2 − β E 2 0 0 E 3 − β E 3 0 0 ) = ( 0 − E ′ 1 − E ′ 2 − E ′ 3 E ′ 1 0 − c B ′ 3 c B ′ 2 E ′ 2 c B ′ 3 0 − c B ′ 1 E ′ 3 − c B ′ 2 c B ′ 1 0 ) (3.562) F ′ ≃ 0 − E 1 − E 2 − E 3 E 1 0 β E 2 β E 3 E 2 − β E 2 0 0 E 3 − β E 3 0 0 = 0 − E ′ 1 − E ′ 2 − E ′ 3 E ′ 1 0 − c B ′ 3 c B ′ 2 E ′ 2 c B ′ 3 0 − c B ′ 1 E ′ 3 − c B ′ 2 c B ′ 1 0 {:(3.562)F^(')≃([0,-E^(1),-E^(2),-E^(3)],[E^(1),0,betaE^(2),betaE^(3)],[E^(2),-betaE^(2),0,0],[E^(3),-betaE^(3),0,0])=([0,-E^('1),-E^('2),-E^('3)],[E^('1),0,-cB^('3),cB^('2)],[E^('2),cB^('3),0,-cB^('1)],[E^('3),-cB^('2),cB^('1),0]):}F^{\prime} \simeq\left(\begin{array}{cccc} 0 & -E^{1} & -E^{2} & -E^{3} \tag{3.562}\\ E^{1} & 0 & \beta E^{2} & \beta E^{3} \\ E^{2} & -\beta E^{2} & 0 & 0 \\ E^{3} & -\beta E^{3} & 0 & 0 \end{array}\right)=\left(\begin{array}{cccc} 0 & -E^{\prime 1} & -E^{\prime 2} & -E^{\prime 3} \\ E^{\prime 1} & 0 & -c B^{\prime 3} & c B^{\prime 2} \\ E^{\prime 2} & c B^{\prime 3} & 0 & -c B^{\prime 1} \\ E^{\prime 3} & -c B^{\prime 2} & c B^{\prime 1} & 0 \end{array}\right)(3.562)F′≃(0−E1−E2−E3E10βE2βE3E2−βE200E3−βE300)=(0−E′1−E′2−E′3E′10−cB′3cB′2E′2cB′30−cB′1E′3−cB′2cB′10)
to linear order in β β beta\betaβ. Thus, the magnetic field in K ′ K ′ K^(')K^{\prime}K′ is
(3.563) B ′ = ( B ′ 1 , B ′ 2 , B ′ 3 ) = v c 2 ( 0 , E 3 , − E 2 ) (3.563) B ′ = B ′ 1 , B ′ 2 , B ′ 3 = v c 2 0 , E 3 , − E 2 {:(3.563)B^(')=(B^('1),B^('2),B^('3))=(v)/(c^(2))(0,E^(3),-E^(2)):}\begin{equation*} \mathbf{B}^{\prime}=\left(B^{\prime 1}, B^{\prime 2}, B^{\prime 3}\right)=\frac{v}{c^{2}}\left(0, E^{3},-E^{2}\right) \tag{3.563} \end{equation*}(3.563)B′=(B′1,B′2,B′3)=vc2(0,E3,−E2)
In addition, the electric field in K ′ K ′ K^(')K^{\prime}K′ is E ′ = ( E ′ 1 , E ′ 2 , E ′ 3 ) = ( E 1 , E 2 , E 3 ) = E E ′ = E ′ 1 , E ′ 2 , E ′ 3 = E 1 , E 2 , E 3 = E E^(')=(E^('1),E^('2),E^('3))=(E^(1),E^(2),E^(3))=E\mathbf{E}^{\prime}=\left(E^{\prime 1}, E^{\prime 2}, E^{\prime 3}\right)=\left(E^{1}, E^{2}, E^{3}\right)=\mathbf{E}E′=(E′1,E′2,E′3)=(E1,E2,E3)=E. Clearly, B ′ B ′ B^(')\mathbf{B}^{\prime}B′ is perpendicular to both the x 1 x 1 x^(1)x^{1}x1-axis (i.e., the direction of the velocity v = − v e 1 v = − v e 1 v=-ve_(1)\mathbf{v}=-v \mathbf{e}_{1}v=−ve1 of K K KKK in K ′ K ′ K^(')K^{\prime}K′ ) and E ′ E ′ E^(')\mathbf{E}^{\prime}E′, since it holds that
(3.564) B ′ ⋅ e 1 = v c 2 ( 0 , E 3 , − E 2 ) ⋅ ( 1 , 0 , 0 ) = 0 (3.565) B ′ ⋅ E = v c 2 ( 0 , E 3 , − E 2 ) ⋅ ( E 1 , E 2 , E 3 ) = 0 (3.564) B ′ ⋅ e 1 = v c 2 0 , E 3 , − E 2 ⋅ ( 1 , 0 , 0 ) = 0 (3.565) B ′ ⋅ E = v c 2 0 , E 3 , − E 2 ⋅ E 1 , E 2 , E 3 = 0 {:[(3.564)B^(')*e_(1)=(v)/(c^(2))(0,E^(3),-E^(2))*(1","0","0)=0],[(3.565)B^(')*E=(v)/(c^(2))(0,E^(3),-E^(2))*(E^(1),E^(2),E^(3))=0]:}\begin{align*} & \mathbf{B}^{\prime} \cdot \mathbf{e}_{1}=\frac{v}{c^{2}}\left(0, E^{3},-E^{2}\right) \cdot(1,0,0)=0 \tag{3.564}\\ & \mathbf{B}^{\prime} \cdot \mathbf{E}=\frac{v}{c^{2}}\left(0, E^{3},-E^{2}\right) \cdot\left(E^{1}, E^{2}, E^{3}\right)=0 \tag{3.565} \end{align*}(3.564)B′⋅e1=vc2(0,E3,−E2)⋅(1,0,0)=0(3.565)B′⋅E=vc2(0,E3,−E2)⋅(E1,E2,E3)=0
Thus, one finds that the magnetic field in K ′ K ′ K^(')K^{\prime}K′ for small velocities is
(3.566) B ′ ≃ v c 2 ( 0 , E 3 , − E 2 ) = − 1 c 2 ( v × E ) (3.566) B ′ ≃ v c 2 0 , E 3 , − E 2 = − 1 c 2 ( v × E ) {:(3.566)B^(')≃(v)/(c^(2))(0,E^(3),-E^(2))=-(1)/(c^(2))(vxxE):}\begin{equation*} \mathbf{B}^{\prime} \simeq \frac{v}{c^{2}}\left(0, E^{3},-E^{2}\right)=-\frac{1}{c^{2}}(\mathbf{v} \times \mathbf{E}) \tag{3.566} \end{equation*}(3.566)B′≃vc2(0,E3,−E2)=−1c2(v×E)
It holds that B ′ ⋅ E ∝ ( v × E ) ⋅ E = 0 B ′ ⋅ E ∝ ( v × E ) ⋅ E = 0 B^(')*Eprop(vxxE)*E=0\mathbf{B}^{\prime} \cdot \mathbf{E} \propto(\mathbf{v} \times \mathbf{E}) \cdot \mathbf{E}=0B′⋅E∝(v×E)⋅E=0 and B ′ ⋅ v ∝ ( v × E ) ⋅ v = 0 B ′ ⋅ v ∝ ( v × E ) ⋅ v = 0 B^(')*vprop(vxxE)*v=0\mathbf{B}^{\prime} \cdot \mathbf{v} \propto(\mathbf{v} \times \mathbf{E}) \cdot \mathbf{v}=0B′⋅v∝(v×E)⋅v=0.

1.116

Consider the electric and magnetic fields in a coordinate system of K K KKK, i.e., E = E = E=\mathbf{E}=E= ( 0 , 1 , 0 ) ( 0 , 1 , 0 ) (0,1,0)(0,1,0)(0,1,0) and B = 0 B = 0 B=0\mathbf{B}=\mathbf{0}B=0, which means that the electromagnetic field components are
(3.567) E 1 = 0 , E 2 = 1 , E 3 = 0 , B 1 = 0 , B 2 = 0 , B 3 = 0 (3.567) E 1 = 0 , E 2 = 1 , E 3 = 0 , B 1 = 0 , B 2 = 0 , B 3 = 0 {:(3.567)E^(1)=0","quadE^(2)=1","quadE^(3)=0","quadB^(1)=0","quadB^(2)=0","quadB^(3)=0:}\begin{equation*} E^{1}=0, \quad E^{2}=1, \quad E^{3}=0, \quad B^{1}=0, \quad B^{2}=0, \quad B^{3}=0 \tag{3.567} \end{equation*}(3.567)E1=0,E2=1,E3=0,B1=0,B2=0,B3=0
Inserting the electromagnetic field components in the coordinate system of K K KKK into the Lorentz transformation corresponding to a velocity boost v x v x v_(x)v_{x}vx in the x 1 x 1 x^(1)x^{1}x1-direction, namely
(3.568) E ′ 1 = E 1 (3.569) E ′ 2 = γ ( v x ) ( E 2 − v x B 3 ) (3.570) E ′ 3 = γ ( v x ) ( E 3 + v x B 2 ) (3.571) B ′ 1 = B 1 (3.572) B ′ 2 = γ ( v x ) ( B 2 + v x c 2 E 3 ) (3.573) B ′ 3 = γ ( v x ) ( B 3 − v x c 2 E 2 ) (3.568) E ′ 1 = E 1 (3.569) E ′ 2 = γ v x E 2 − v x B 3 (3.570) E ′ 3 = γ v x E 3 + v x B 2 (3.571) B ′ 1 = B 1 (3.572) B ′ 2 = γ v x B 2 + v x c 2 E 3 (3.573) B ′ 3 = γ v x B 3 − v x c 2 E 2 {:[(3.568)E^('1)=E^(1)],[(3.569)E^('2)=gamma(v_(x))(E^(2)-v_(x)B^(3))],[(3.570)E^('3)=gamma(v_(x))(E^(3)+v_(x)B^(2))],[(3.571)B^('1)=B^(1)],[(3.572)B^('2)=gamma(v_(x))(B^(2)+(v_(x))/(c^(2))E^(3))],[(3.573)B^('3)=gamma(v_(x))(B^(3)-(v_(x))/(c^(2))E^(2))]:}\begin{align*} & E^{\prime 1}=E^{1} \tag{3.568}\\ & E^{\prime 2}=\gamma\left(v_{x}\right)\left(E^{2}-v_{x} B^{3}\right) \tag{3.569}\\ & E^{\prime 3}=\gamma\left(v_{x}\right)\left(E^{3}+v_{x} B^{2}\right) \tag{3.570}\\ & B^{\prime 1}=B^{1} \tag{3.571}\\ & B^{\prime 2}=\gamma\left(v_{x}\right)\left(B^{2}+\frac{v_{x}}{c^{2}} E^{3}\right) \tag{3.572}\\ & B^{\prime 3}=\gamma\left(v_{x}\right)\left(B^{3}-\frac{v_{x}}{c^{2}} E^{2}\right) \tag{3.573} \end{align*}(3.568)E′1=E1(3.569)E′2=γ(vx)(E2−vxB3)(3.570)E′3=γ(vx)(E3+vxB2)(3.571)B′1=B1(3.572)B′2=γ(vx)(B2+vxc2E3)(3.573)B′3=γ(vx)(B3−vxc2E2)
the electromagnetic field components in the coordinate system of K ′ K ′ K^(')K^{\prime}K′ are
(3.574) E ′ 1 = 0 , E ′ 2 = γ ( v x ) , E ′ 3 = 0 , B ′ 1 = 0 , B ′ 2 = 0 , B ′ 3 = − v x c 2 γ ( v x ) , (3.574) E ′ 1 = 0 , E ′ 2 = γ v x , E ′ 3 = 0 , B ′ 1 = 0 , B ′ 2 = 0 , B ′ 3 = − v x c 2 γ v x , {:(3.574)E^('1)=0","quadE^('2)=gamma(v_(x))","quadE^('3)=0","quadB^('1)=0","quadB^('2)=0","quadB^('3)=-(v_(x))/(c^(2))gamma(v_(x))",":}\begin{equation*} E^{\prime 1}=0, \quad E^{\prime 2}=\gamma\left(v_{x}\right), \quad E^{\prime 3}=0, \quad B^{\prime 1}=0, \quad B^{\prime 2}=0, \quad B^{\prime 3}=-\frac{v_{x}}{c^{2}} \gamma\left(v_{x}\right), \tag{3.574} \end{equation*}(3.574)E′1=0,E′2=γ(vx),E′3=0,B′1=0,B′2=0,B′3=−vxc2γ(vx),
where γ ( v ) ≡ 1 1 − v 2 / c 2 γ ( v ) ≡ 1 1 − v 2 / c 2 gamma(v)-=(1)/(sqrt(1-v^(2)//c^(2)))\gamma(v) \equiv \frac{1}{\sqrt{1-v^{2} / c^{2}}}γ(v)≡11−v2/c2. Similarly, with a velocity boost v y v y v_(y)v_{y}vy in the x ′ 2 x ′ 2 x^('2)x^{\prime 2}x′2-direction, the electromagnetic field components in the coordinate system of K ′ ′ K ′ ′ K^('')K^{\prime \prime}K′′ are given by
(3.575) E ′ ′ 1 = − v x v y c 2 γ ( v x ) γ ( v y ) (3.576) E ′ ′ 2 = γ ( v x ) (3.577) E ′ ′ 3 = 0 (3.578) B ′ ′ 1 = 0 (3.579) B ′ ′ 2 = 0 (3.580) B ′ ′ 3 = − v x c 2 γ ( v x ) γ ( v y ) (3.575) E ′ ′ 1 = − v x v y c 2 γ v x γ v y (3.576) E ′ ′ 2 = γ v x (3.577) E ′ ′ 3 = 0 (3.578) B ′ ′ 1 = 0 (3.579) B ′ ′ 2 = 0 (3.580) B ′ ′ 3 = − v x c 2 γ v x γ v y {:[(3.575)E^(''1)=-(v_(x)v_(y))/(c^(2))gamma(v_(x))gamma(v_(y))],[(3.576)E^(''2)=gamma(v_(x))],[(3.577)E^(''3)=0],[(3.578)B^(''1)=0],[(3.579)B^(''2)=0],[(3.580)B^(''3)=-(v_(x))/(c^(2))gamma(v_(x))gamma(v_(y))]:}\begin{align*} & E^{\prime \prime 1}=-\frac{v_{x} v_{y}}{c^{2}} \gamma\left(v_{x}\right) \gamma\left(v_{y}\right) \tag{3.575}\\ & E^{\prime \prime 2}=\gamma\left(v_{x}\right) \tag{3.576}\\ & E^{\prime \prime 3}=0 \tag{3.577}\\ & B^{\prime \prime 1}=0 \tag{3.578}\\ & B^{\prime \prime 2}=0 \tag{3.579}\\ & B^{\prime \prime 3}=-\frac{v_{x}}{c^{2}} \gamma\left(v_{x}\right) \gamma\left(v_{y}\right) \tag{3.580} \end{align*}(3.575)E′′1=−vxvyc2γ(vx)γ(vy)(3.576)E′′2=γ(vx)(3.577)E′′3=0(3.578)B′′1=0(3.579)B′′2=0(3.580)B′′3=−vxc2γ(vx)γ(vy)

1.117

The electric field E E E\boldsymbol{E}E due to a point charge q q qqq at the origin is known to be
(3.581) E ( x ) = q x 4 π ϵ 0 r 3 . (3.581) E ( x ) = q x 4 π ϵ 0 r 3 . {:(3.581)E(x)=(qx)/(4piepsilon_(0)r^(3)).:}\begin{equation*} \boldsymbol{E}(x)=\frac{q \boldsymbol{x}}{4 \pi \epsilon_{0} r^{3}} . \tag{3.581} \end{equation*}(3.581)E(x)=qx4πϵ0r3.
After giving an observer a boost to velocity v v vvv along the positive x 1 x 1 x^(1)x^{1}x1-axis, the field is transformed to
(3.582) E ′ ( x ′ ) = q 4 π ϵ 0 r 3 ( x 1 , x 2 cosh ⁡ θ , x 3 cosh ⁡ θ ) , (3.583) c B ′ ( x ′ ) = q 4 π ϵ 0 r 3 ( 0 , x 3 sinh ⁡ θ , − x 2 sinh ⁡ θ ) , (3.582) E ′ x ′ = q 4 π ϵ 0 r 3 x 1 , x 2 cosh ⁡ θ , x 3 cosh ⁡ θ , (3.583) c B ′ x ′ = q 4 π ϵ 0 r 3 0 , x 3 sinh ⁡ θ , − x 2 sinh ⁡ θ , {:[(3.582)E^(')(x^('))=(q)/(4piepsilon_(0)r^(3))(x^(1),x^(2)cosh theta,x^(3)cosh theta)","],[(3.583)cB^(')(x^('))=(q)/(4piepsilon_(0)r^(3))(0,x^(3)sinh theta,-x^(2)sinh theta)","]:}\begin{align*} \boldsymbol{E}^{\prime}\left(x^{\prime}\right) & =\frac{q}{4 \pi \epsilon_{0} r^{3}}\left(x^{1}, x^{2} \cosh \theta, x^{3} \cosh \theta\right), \tag{3.582}\\ c \boldsymbol{B}^{\prime}\left(x^{\prime}\right) & =\frac{q}{4 \pi \epsilon_{0} r^{3}}\left(0, x^{3} \sinh \theta,-x^{2} \sinh \theta\right), \tag{3.583} \end{align*}(3.582)E′(x′)=q4πϵ0r3(x1,x2cosh⁡θ,x3cosh⁡θ),(3.583)cB′(x′)=q4πϵ0r3(0,x3sinh⁡θ,−x2sinh⁡θ),
where cosh ⁡ θ = γ ( v ) ≡ 1 / 1 − v 2 / c 2 , sinh ⁡ θ = v γ ( v ) / c , cosh ⁡ θ = γ ( v ) ≡ 1 / 1 − v 2 / c 2 , sinh ⁡ θ = v γ ( v ) / c , quad cosh theta=gamma(v)-=1//sqrt(1-v^(2)//c^(2)),quad sinh theta=v gamma(v)//c,quad\quad \cosh \theta=\gamma(v) \equiv 1 / \sqrt{1-v^{2} / c^{2}}, \quad \sinh \theta=v \gamma(v) / c, \quadcosh⁡θ=γ(v)≡1/1−v2/c2,sinh⁡θ=vγ(v)/c, and r = r = quad r=\quad r=r= ( x 1 ) 2 + ( x 2 ) 2 + ( x 3 ) 2 x 1 2 + x 2 2 + x 3 2 sqrt((x^(1))^(2)+(x^(2))^(2)+(x^(3))^(2))\sqrt{\left(x^{1}\right)^{2}+\left(x^{2}\right)^{2}+\left(x^{3}\right)^{2}}(x1)2+(x2)2+(x3)2.
Note that in order to compute the electromagnetic field strengths at the point x ′ x ′ x^(')x^{\prime}x′, we have to write the x x xxx-coordinates on the right-hand side of the equations in terms of the new x ′ x ′ x^(')x^{\prime}x′-coordinates. After doing so, we obtain
(3.584) E ′ ( x ′ ) = q 4 π ϵ 0 r 3 ( x ′ 1 + v t ′ , x ′ 2 , x ′ 3 ) cosh ⁡ θ (3.585) c B ′ ( x ′ ) = q 4 π ϵ 0 r 3 ( 0 , x ′ 3 , − x ′ 2 ) sinh ⁡ θ (3.584) E ′ x ′ = q 4 π ϵ 0 r 3 x ′ 1 + v t ′ , x ′ 2 , x ′ 3 cosh ⁡ θ (3.585) c B ′ x ′ = q 4 π ϵ 0 r 3 0 , x ′ 3 , − x ′ 2 sinh ⁡ θ {:[(3.584)E^(')(x^('))=(q)/(4piepsilon_(0)r^(3))(x^('1)+vt^('),x^('2),x^('3))cosh theta],[(3.585)cB^(')(x^('))=(q)/(4piepsilon_(0)r^(3))(0,x^('3),-x^('2))sinh theta]:}\begin{align*} \boldsymbol{E}^{\prime}\left(x^{\prime}\right) & =\frac{q}{4 \pi \epsilon_{0} r^{3}}\left(x^{\prime 1}+v t^{\prime}, x^{\prime 2}, x^{\prime 3}\right) \cosh \theta \tag{3.584}\\ c \boldsymbol{B}^{\prime}\left(x^{\prime}\right) & =\frac{q}{4 \pi \epsilon_{0} r^{3}}\left(0, x^{\prime 3},-x^{\prime 2}\right) \sinh \theta \tag{3.585} \end{align*}(3.584)E′(x′)=q4πϵ0r3(x′1+vt′,x′2,x′3)cosh⁡θ(3.585)cB′(x′)=q4πϵ0r3(0,x′3,−x′2)sinh⁡θ
where r = cosh 2 ⁡ θ ( x ′ 1 + v t ′ ) 2 + ( x ′ 2 ) 2 + ( x ′ 3 ) 2 r = cosh 2 ⁡ θ x ′ 1 + v t ′ 2 + x ′ 2 2 + x ′ 3 2 r=sqrt(cosh^(2)theta(x^('1)+vt^('))^(2)+(x^('2))^(2)+(x^('3))^(2))r=\sqrt{\cosh ^{2} \theta\left(x^{\prime 1}+v t^{\prime}\right)^{2}+\left(x^{\prime 2}\right)^{2}+\left(x^{\prime 3}\right)^{2}}r=cosh2⁡θ(x′1+vt′)2+(x′2)2+(x′3)2.
For small velocities, i.e., v ≪ c v ≪ c v≪cv \ll cv≪c, we recover the classical formulas
(3.586) E ′ 1 ( x ′ ) = q 4 π ϵ 0 ( x ′ 1 + v t ′ ) [ ( x ′ 1 + v t ′ ) 2 + ( x ′ 2 ) 2 + ( x ′ 3 ) 2 ] − 3 / 2 (3.587) E ′ 2 ( x ′ ) = q 4 π ϵ 0 x ′ 2 [ ( x ′ 1 + v t ′ ) 2 + ( x ′ 2 ) 2 + ( x ′ 3 ) 2 ] − 3 / 2 (3.588) E ′ 3 ( x ′ ) = q 4 π ϵ 0 x ′ 3 [ ( x ′ 1 + v t ′ ) 2 + ( x ′ 2 ) 2 + ( x ′ 3 ) 2 ] − 3 / 2 (3.589) B ′ 1 ( x ′ ) = 0 (3.590) B ′ 2 ( x ′ ) = q 4 π ϵ 0 x ′ 3 v c 2 [ ( x ′ 1 + v t ′ ) 2 + ( x ′ 2 ) 2 + ( x ′ 3 ) 2 ] − 3 / 2 (3.591) B ′ 3 ( x ′ ) = − q 4 π ϵ 0 x ′ 2 v c 2 [ ( x ′ 1 + v t ′ ) 2 + ( x ′ 2 ) 2 + ( x ′ 3 ) 2 ] − 3 / 2 (3.586) E ′ 1 x ′ = q 4 π ϵ 0 x ′ 1 + v t ′ x ′ 1 + v t ′ 2 + x ′ 2 2 + x ′ 3 2 − 3 / 2 (3.587) E ′ 2 x ′ = q 4 π ϵ 0 x ′ 2 x ′ 1 + v t ′ 2 + x ′ 2 2 + x ′ 3 2 − 3 / 2 (3.588) E ′ 3 x ′ = q 4 π ϵ 0 x ′ 3 x ′ 1 + v t ′ 2 + x ′ 2 2 + x ′ 3 2 − 3 / 2 (3.589) B ′ 1 x ′ = 0 (3.590) B ′ 2 x ′ = q 4 π ϵ 0 x ′ 3 v c 2 x ′ 1 + v t ′ 2 + x ′ 2 2 + x ′ 3 2 − 3 / 2 (3.591) B ′ 3 x ′ = − q 4 π ϵ 0 x ′ 2 v c 2 x ′ 1 + v t ′ 2 + x ′ 2 2 + x ′ 3 2 − 3 / 2 {:[(3.586)E^('1)(x^('))=(q)/(4piepsilon_(0))(x^('1)+vt^('))[(x^('1)+vt^('))^(2)+(x^('2))^(2)+(x^('3))^(2)]^(-3//2)],[(3.587)E^('2)(x^('))=(q)/(4piepsilon_(0))x^('2)[(x^('1)+vt^('))^(2)+(x^('2))^(2)+(x^('3))^(2)]^(-3//2)],[(3.588)E^('3)(x^('))=(q)/(4piepsilon_(0))x^('3)[(x^('1)+vt^('))^(2)+(x^('2))^(2)+(x^('3))^(2)]^(-3//2)],[(3.589)B^('1)(x^('))=0],[(3.590)B^('2)(x^('))=(q)/(4piepsilon_(0))x^('3)(v)/(c^(2))[(x^('1)+vt^('))^(2)+(x^('2))^(2)+(x^('3))^(2)]^(-3//2)],[(3.591)B^('3)(x^('))=-(q)/(4piepsilon_(0))x^('2)(v)/(c^(2))[(x^('1)+vt^('))^(2)+(x^('2))^(2)+(x^('3))^(2)]^(-3//2)]:}\begin{align*} & E^{\prime 1}\left(x^{\prime}\right)=\frac{q}{4 \pi \epsilon_{0}}\left(x^{\prime 1}+v t^{\prime}\right)\left[\left(x^{\prime 1}+v t^{\prime}\right)^{2}+\left(x^{\prime 2}\right)^{2}+\left(x^{\prime 3}\right)^{2}\right]^{-3 / 2} \tag{3.586}\\ & E^{\prime 2}\left(x^{\prime}\right)=\frac{q}{4 \pi \epsilon_{0}} x^{\prime 2}\left[\left(x^{\prime 1}+v t^{\prime}\right)^{2}+\left(x^{\prime 2}\right)^{2}+\left(x^{\prime 3}\right)^{2}\right]^{-3 / 2} \tag{3.587}\\ & E^{\prime 3}\left(x^{\prime}\right)=\frac{q}{4 \pi \epsilon_{0}} x^{\prime 3}\left[\left(x^{\prime 1}+v t^{\prime}\right)^{2}+\left(x^{\prime 2}\right)^{2}+\left(x^{\prime 3}\right)^{2}\right]^{-3 / 2} \tag{3.588}\\ & B^{\prime 1}\left(x^{\prime}\right)=0 \tag{3.589}\\ & B^{\prime 2}\left(x^{\prime}\right)=\frac{q}{4 \pi \epsilon_{0}} x^{\prime 3} \frac{v}{c^{2}}\left[\left(x^{\prime 1}+v t^{\prime}\right)^{2}+\left(x^{\prime 2}\right)^{2}+\left(x^{\prime 3}\right)^{2}\right]^{-3 / 2} \tag{3.590}\\ & B^{\prime 3}\left(x^{\prime}\right)=-\frac{q}{4 \pi \epsilon_{0}} x^{\prime 2} \frac{v}{c^{2}}\left[\left(x^{\prime 1}+v t^{\prime}\right)^{2}+\left(x^{\prime 2}\right)^{2}+\left(x^{\prime 3}\right)^{2}\right]^{-3 / 2} \tag{3.591} \end{align*}(3.586)E′1(x′)=q4πϵ0(x′1+vt′)[(x′1+vt′)2+(x′2)2+(x′3)2]−3/2(3.587)E′2(x′)=q4πϵ0x′2[(x′1+vt′)2+(x′2)2+(x′3)2]−3/2(3.588)E′3(x′)=q4πϵ0x′3[(x′1+vt′)2+(x′2)2+(x′3)2]−3/2(3.589)B′1(x′)=0(3.590)B′2(x′)=q4πϵ0x′3vc2[(x′1+vt′)2+(x′2)2+(x′3)2]−3/2(3.591)B′3(x′)=−q4πϵ0x′2vc2[(x′1+vt′)2+(x′2)2+(x′3)2]−3/2
for the electromagnetic field of a point charge moving along the negative x ′ 1 x ′ 1 x^('1)x^{\prime 1}x′1-axis with velocity v v vvv, i.e., we have an electric current in the negative x ′ 1 x ′ 1 x^('1)x^{\prime 1}x′1-direction.

1.118

Maxwell's equations describe how sources (charges and currents) give rise to electric and magnetic fields, whereas the Lorentz force law describes how the field strengths determine the trajectory of a moving test particle with rest mass m m mmm and electric charge q q qqq. Let us parametrize the trajectory of the particle as x = x ( s ) x = x ( s ) x=x(s)x=x(s)x=x(s), where s s sss is the proper time parameter. The Lorentz force law is then given by
(3.592) m c 2 x ¨ μ ( s ) = q x ˙ v ( s ) F μ v ( x ( s ) ) (3.592) m c 2 x ¨ μ ( s ) = q x ˙ v ( s ) F μ v ( x ( s ) ) {:(3.592)mc^(2)x^(¨)^(mu)(s)=qx^(˙)_(v)(s)F^(mu v)(x(s)):}\begin{equation*} m c^{2} \ddot{x}^{\mu}(s)=q \dot{x}_{v}(s) F^{\mu v}(x(s)) \tag{3.592} \end{equation*}(3.592)mc2x¨μ(s)=qx˙v(s)Fμv(x(s))
which is covariant under Lorentz transformations, i.e., both sides of the equation transform as 4 -vectors, and where F μ ν F μ ν F^(mu nu)F^{\mu \nu}Fμν is the electromagnetic field strength tensor. In order to understand the physical meaning of the Lorentz force law, we will first replace proper time derivatives by ordinary time derivatives, using x 0 = c t x 0 = c t x^(0)=ctx^{0}=c tx0=ct. For the time derivatives of the spatial components of x = x ( s ) x = x ( s ) x=x(s)x=x(s)x=x(s), we have
(3.593) x ˙ = d x d s = d x d t d t d s = u 1 c d x 0 d s = 1 c u x ˙ 0 (3.593) x ˙ = d x d s = d x d t d t d s = u 1 c d x 0 d s = 1 c u x ˙ 0 {:(3.593)x^(˙)=(dx)/(ds)=(dx)/(dt)(dt)/(ds)=u(1)/(c)(dx^(0))/(ds)=(1)/(c)ux^(˙)^(0):}\begin{equation*} \dot{\boldsymbol{x}}=\frac{d \boldsymbol{x}}{d s}=\frac{d x}{d t} \frac{d t}{d s}=\boldsymbol{u} \frac{1}{c} \frac{d x^{0}}{d s}=\frac{1}{c} \boldsymbol{u} \dot{x}^{0} \tag{3.593} \end{equation*}(3.593)x˙=dxds=dxdtdtds=u1cdx0ds=1cux˙0
Then, from the spatial part of the Lorentz force law and using the definitions of the electromagnetic field strengths, we obtain
m c 2 x ¨ i = m c 2 d x ˙ i d s = m c 2 d t d s d d t ( d x i d t ⋅ d t d s ) = m d x 0 d s ⋅ d d t ⋅ ( u i d x 0 d s ) = d x 0 d s ⋅ d d t ( m u i x ˙ 0 ) m c 2 x ¨ i = m c 2 d x ˙ i d s = m c 2 d t d s d d t d x i d t ⋅ d t d s = m d x 0 d s ⋅ d d t ⋅ u i d x 0 d s = d x 0 d s ⋅ d d t m u i x ˙ 0 {:[mc^(2)x^(¨)^(i)=mc^(2)(dx^(˙)^(i))/(ds)=mc^(2)(dt)/(ds)(d)/(dt)((dx^(i))/(dt)*(dt)/(ds))],[=m(dx^(0))/(ds)*(d)/(dt)*(u^(i)(dx^(0))/(ds))=(dx^(0))/(ds)*(d)/(dt)(mu^(i)x^(˙)^(0))]:}\begin{aligned} m c^{2} \ddot{x}^{i} & =m c^{2} \frac{d \dot{x}^{i}}{d s}=m c^{2} \frac{d t}{d s} \frac{d}{d t}\left(\frac{d x^{i}}{d t} \cdot \frac{d t}{d s}\right) \\ & =m \frac{d x^{0}}{d s} \cdot \frac{d}{d t} \cdot\left(u^{i} \frac{d x^{0}}{d s}\right)=\frac{d x^{0}}{d s} \cdot \frac{d}{d t}\left(m u^{i} \dot{x}^{0}\right) \end{aligned}mc2x¨i=mc2dx˙ids=mc2dtdsddt(dxidt⋅dtds)=mdx0ds⋅ddt⋅(uidx0ds)=dx0ds⋅ddt(muix˙0)
= q x ˙ v F i v = q ( x ˙ 0 F i 0 + x ˙ j F i j ) = q ( d x 0 d s E i + d t d s ( u × c B ) i ) (3.594) = q ( d x 0 d s E i + d x 0 d s ⋅ ( u × B ) i ) = d x 0 d s q ( E + u × B ) i . = q x ˙ v F i v = q x ˙ 0 F i 0 + x ˙ j F i j = q d x 0 d s E i + d t d s ( u × c B ) i (3.594) = q d x 0 d s E i + d x 0 d s ⋅ ( u × B ) i = d x 0 d s q ( E + u × B ) i . {:[=qx^(˙)_(v)F^(iv)=q(x^(˙)_(0)F^(i0)+x^(˙)_(j)F^(ij))=q((dx_(0))/(ds)E^(i)+(dt)/(ds)(u xx cB)^(i))],[(3.594)=q((dx^(0))/(ds)E^(i)+(dx^(0))/(ds)*(u xx B)^(i))=(dx^(0))/(ds)q(E+u xx B)^(i).]:}\begin{align*} & =q \dot{x}_{v} F^{i v}=q\left(\dot{x}_{0} F^{i 0}+\dot{x}_{j} F^{i j}\right)=q\left(\frac{d x_{0}}{d s} E^{i}+\frac{d t}{d s}(\boldsymbol{u} \times c \boldsymbol{B})^{i}\right) \\ & =q\left(\frac{d x^{0}}{d s} E^{i}+\frac{d x^{0}}{d s} \cdot(\boldsymbol{u} \times \boldsymbol{B})^{i}\right)=\frac{d x^{0}}{d s} q(\boldsymbol{E}+\boldsymbol{u} \times \boldsymbol{B})^{i} . \tag{3.594} \end{align*}=qx˙vFiv=q(x˙0Fi0+x˙jFij)=q(dx0dsEi+dtds(u×cB)i)(3.594)=q(dx0dsEi+dx0ds⋅(u×B)i)=dx0dsq(E+u×B)i.
Thus, the factors d x 0 d s d x 0 d s (dx^(0))/(ds)\frac{d x^{0}}{d s}dx0ds cancel of both sides of this equation to give
(3.595) d d t ( m u x ˙ 0 ) = q ( E + u × B ) . (3.595) d d t m u x ˙ 0 = q ( E + u × B ) . {:(3.595)(d)/(dt)(mux^(˙)^(0))=q(E+u xx B).:}\begin{equation*} \frac{d}{d t}\left(m \boldsymbol{u} \dot{x}^{0}\right)=q(\boldsymbol{E}+\boldsymbol{u} \times \boldsymbol{B}) . \tag{3.595} \end{equation*}(3.595)ddt(mux˙0)=q(E+u×B).
Now, what is m u x ˙ 0 m u x ˙ 0 mux^(˙)^(0)m u \dot{x}^{0}mux˙0 ? The relativistic 4-momentum is defined as p μ = m c x ˙ μ p μ = m c x ˙ μ p^(mu)=mcx^(˙)^(mu)p^{\mu}=m c \dot{x}^{\mu}pμ=mcx˙μ, so we have p = m c x ˙ = m u x ˙ 0 p = m c x ˙ = m u x ˙ 0 p=mcx^(˙)=mux^(˙)^(0)\boldsymbol{p}=m c \dot{\boldsymbol{x}}=m \boldsymbol{u} \dot{x}^{0}p=mcx˙=mux˙0 ( note that p ≠ m u ) p ≠ m u {:p!=mu)\left.\boldsymbol{p} \neq m \boldsymbol{u}\right)p≠mu), using x ˙ = 1 c u x ˙ 0 x ˙ = 1 c u x ˙ 0 x^(˙)=(1)/(c)ux^(˙)^(0)\dot{\boldsymbol{x}}=\frac{1}{c} \boldsymbol{u} \dot{x}^{0}x˙=1cux˙0. Thus, the Lorentz force law for the spatial components is given by
(3.596) d p d t = q ( E + u × B ) , (3.596) d p d t = q ( E + u × B ) , {:(3.596)(dp)/(dt)=q(E+u xx B)",":}\begin{equation*} \frac{d p}{d t}=q(\boldsymbol{E}+\boldsymbol{u} \times \boldsymbol{B}), \tag{3.596} \end{equation*}(3.596)dpdt=q(E+u×B),
where p p ppp is the relativistic 3-momentum and E E E\boldsymbol{E}E and B B B\boldsymbol{B}B are the electric and magnetic fields, respectively. Furthermore, what is x ˙ 0 x ˙ 0 x^(˙)^(0)\dot{x}^{0}x˙0 ? The proper time parameter s s sss is defined such that x ˙ 2 = ( x ˙ 0 ) 2 − x ˙ 2 = 1 x ˙ 2 = x ˙ 0 2 − x ˙ 2 = 1 x^(˙)^(2)=(x^(˙)^(0))^(2)-x^(˙)^(2)=1\dot{x}^{2}=\left(\dot{x}^{0}\right)^{2}-\dot{\boldsymbol{x}}^{2}=1x˙2=(x˙0)2−x˙2=1. Combining this with c x ˙ = u x ˙ 0 c x ˙ = u x ˙ 0 cx^(˙)=ux^(˙)^(0)c \dot{x}=u \dot{x}^{0}cx˙=ux˙0, we deduce
(3.597) u = | u | = c x ˙ 0 | x ˙ | = c | x ˙ | 1 + | x ˙ | 2 ⇒ | x ˙ | = u / c 1 − u 2 / c 2 . (3.597) u = | u | = c x ˙ 0 | x ˙ | = c | x ˙ | 1 + | x ˙ | 2 ⇒ | x ˙ | = u / c 1 − u 2 / c 2 . {:(3.597)u=|u|=(c)/(x^(˙)^(0))|x^(˙)|=(c|(x^(˙))|)/(sqrt(1+|x^(˙)|^(2)))=>|x^(˙)|=(u//c)/(sqrt(1-u^(2)//c^(2))).:}\begin{equation*} u=|\boldsymbol{u}|=\frac{c}{\dot{x}^{0}}|\dot{\boldsymbol{x}}|=\frac{c|\dot{\boldsymbol{x}}|}{\sqrt{1+|\dot{\boldsymbol{x}}|^{2}}} \Rightarrow|\dot{\boldsymbol{x}}|=\frac{u / c}{\sqrt{1-u^{2} / c^{2}}} . \tag{3.597} \end{equation*}(3.597)u=|u|=cx˙0|x˙|=c|x˙|1+|x˙|2⇒|x˙|=u/c1−u2/c2.
Thus, we have
(3.598) x ˙ 0 = 1 + x ˙ 2 = 1 1 − u 2 / c 2 ≡ γ ( u ) . (3.598) x ˙ 0 = 1 + x ˙ 2 = 1 1 − u 2 / c 2 ≡ γ ( u ) . {:(3.598)x^(˙)^(0)=sqrt(1+x^(˙)^(2))=(1)/(sqrt(1-u^(2)//c^(2)))-=gamma(u).:}\begin{equation*} \dot{x}^{0}=\sqrt{1+\dot{\boldsymbol{x}}^{2}}=\frac{1}{\sqrt{1-u^{2} / c^{2}}} \equiv \gamma(u) . \tag{3.598} \end{equation*}(3.598)x˙0=1+x˙2=11−u2/c2≡γ(u).
Assuming E E E\boldsymbol{E}E is a constant electric field and B = 0 B = 0 B=0\boldsymbol{B}=\mathbf{0}B=0 as well as multiplying both sides with d r d r drd \boldsymbol{r}dr, we find that
(3.599) d p d t ⋅ d r = q E ⋅ d r = q E ⋅ u d t (3.599) d p d t ⋅ d r = q E ⋅ d r = q E ⋅ u d t {:(3.599)(dp)/(dt)*dr=qE*dr=qE*udt:}\begin{equation*} \frac{d p}{d t} \cdot d \boldsymbol{r}=q \boldsymbol{E} \cdot d \boldsymbol{r}=q \boldsymbol{E} \cdot \boldsymbol{u} d t \tag{3.599} \end{equation*}(3.599)dpdt⋅dr=qE⋅dr=qE⋅udt
The left-hand side is then
(3.600) d p d t ⋅ d r = d d t ( m u x ˙ 0 ) ⋅ u d t = d ( m u x ˙ 0 ) ⋅ u = m u ⋅ d ( u x ˙ 0 ) (3.600) d p d t ⋅ d r = d d t m u x ˙ 0 ⋅ u d t = d m u x ˙ 0 ⋅ u = m u ⋅ d u x ˙ 0 {:(3.600)(dp)/(dt)*dr=(d)/(dt)(mux^(˙)^(0))*udt=d(mux^(˙)^(0))*u=mu*d(ux^(˙)^(0)):}\begin{equation*} \frac{d \boldsymbol{p}}{d t} \cdot d \boldsymbol{r}=\frac{d}{d t}\left(m \boldsymbol{u} \dot{x}^{0}\right) \cdot \boldsymbol{u} d t=d\left(m \boldsymbol{u} \dot{x}^{0}\right) \cdot \boldsymbol{u}=m \boldsymbol{u} \cdot d\left(\boldsymbol{u} \dot{x}^{0}\right) \tag{3.600} \end{equation*}(3.600)dpdt⋅dr=ddt(mux˙0)⋅udt=d(mux˙0)⋅u=mu⋅d(ux˙0)
Inserting x ˙ 0 = 1 / 1 − u 2 / c 2 x ˙ 0 = 1 / 1 − u 2 / c 2 x^(˙)^(0)=1//sqrt(1-u^(2)//c^(2))\dot{x}^{0}=1 / \sqrt{1-u^{2} / c^{2}}x˙0=1/1−u2/c2 and using straightforward differential calculus, gives for the left-hand side
(3.601) d p d t ⋅ d r = m u ⋅ d ( u 1 − u 2 / c 2 ) = m u ⋅ d u ( 1 − u 2 / c 2 ) 3 / 2 = d ( m c 2 1 − u 2 / c 2 ) (3.601) d p d t ⋅ d r = m u ⋅ d u 1 − u 2 / c 2 = m u ⋅ d u 1 − u 2 / c 2 3 / 2 = d m c 2 1 − u 2 / c 2 {:(3.601)(dp)/(dt)*dr=mu*d((u)/(sqrt(1-u^(2)//c^(2))))=(mu*du)/((1-u^(2)//c^(2))^(3//2))=d((mc^(2))/(sqrt(1-u^(2)//c^(2)))):}\begin{equation*} \frac{d p}{d t} \cdot d \boldsymbol{r}=m \boldsymbol{u} \cdot d\left(\frac{\boldsymbol{u}}{\sqrt{1-u^{2} / c^{2}}}\right)=\frac{m \boldsymbol{u} \cdot d \boldsymbol{u}}{\left(1-u^{2} / c^{2}\right)^{3 / 2}}=d\left(\frac{m c^{2}}{\sqrt{1-u^{2} / c^{2}}}\right) \tag{3.601} \end{equation*}(3.601)dpdt⋅dr=mu⋅d(u1−u2/c2)=mu⋅du(1−u2/c2)3/2=d(mc21−u2/c2)
Integrating both sides from the origin, where u = 0 u = 0 u=0u=0u=0, to the displacement r r rrr, where the velocity momentarily is u u uuu, we find that
∫ u ( 0 ) = 0 u ( r ) d p d t ⋅ d r = ∫ r = 0 r q E ⋅ d r ⇒ m c 2 1 − u 2 / c 2 − m c 2 = q E r (3.602) ⇒ 1 1 − u 2 / c 2 = 1 + q E r m c 2 ∫ u ( 0 ) = 0 u ( r )   d p d t ⋅ d r = ∫ r = 0 r   q E ⋅ d r ⇒ m c 2 1 − u 2 / c 2 − m c 2 = q E r (3.602) ⇒ 1 1 − u 2 / c 2 = 1 + q E r m c 2 {:[int_(u(0)=0)^(u(r))(dp)/(dt)*dr=int_(r=0)^(r)qE*dr=>(mc^(2))/(sqrt(1-u^(2)//c^(2)))-mc^(2)=qEr],[(3.602)=>(1)/(sqrt(1-u^(2)//c^(2)))=1+(qEr)/(mc^(2))]:}\begin{align*} \int_{u(0)=0}^{u(r)} \frac{d p}{d t} \cdot d \boldsymbol{r}=\int_{r=0}^{r} q \boldsymbol{E} \cdot d \boldsymbol{r} & \Rightarrow \frac{m c^{2}}{\sqrt{1-u^{2} / c^{2}}}-m c^{2}=q E r \\ & \Rightarrow \frac{1}{\sqrt{1-u^{2} / c^{2}}}=1+\frac{q E r}{m c^{2}} \tag{3.602} \end{align*}∫u(0)=0u(r)dpdt⋅dr=∫r=0rqE⋅dr⇒mc21−u2/c2−mc2=qEr(3.602)⇒11−u2/c2=1+qErmc2
since the electric field E E E\boldsymbol{E}E is constant along the trajectory. Finally, introducing x ≡ x ≡ x-=x \equivx≡ 1 + q E r m c 2 1 + q E r m c 2 1+(qEr)/(mc^(2))1+\frac{q E r}{m c^{2}}1+qErmc2 and solving for u u uuu, we obtain
(3.603) u ( r ) = c 1 − x − 2 = c 1 − ( 1 + q E r m c 2 ) − 2 (3.603) u ( r ) = c 1 − x − 2 = c 1 − 1 + q E r m c 2 − 2 {:(3.603)u(r)=csqrt(1-x^(-2))=csqrt(1-(1+(qEr)/(mc^(2)))^(-2)):}\begin{equation*} u(r)=c \sqrt{1-x^{-2}}=c \sqrt{1-\left(1+\frac{q E r}{m c^{2}}\right)^{-2}} \tag{3.603} \end{equation*}(3.603)u(r)=c1−x−2=c1−(1+qErmc2)−2
which is the velocity u ( r ) u ( r ) u(r)u(r)u(r) of the particle as a function of the displacement r r rrr from the origin along the direction of motion.
An alternative solution is to note that the electric field in the momentary rest frame K ∘ K ∘ K^(@)\stackrel{\circ}{K}K∘ of the electron is, according to the transformation equations for the field tensor, equal to the electric field in the laboratory system. Using units with c = 1 c = 1 c=1c=1c=1, the acceleration of the electron relative to K ∘ K ∘ K^(@)\stackrel{\circ}{K}K∘ is therefore a ∘ = e E / m 0 a ∘ = e E / m 0 a^(@)=eE//m_(0)\stackrel{\circ}{a}=e E / m_{0}a∘=eE/m0, where e e eee is the electron charge and m 0 m 0 m_(0)m_{0}m0 is the electron rest mass. We obtain the answer by changing g → e E / m 0 g → e E / m 0 g rarr eE//m_(0)g \rightarrow e E / m_{0}g→eE/m0 in the formula [see Problem 1.41, Eq. (3.166)]
(3.604) x = 1 g [ 1 + ( g t ) 2 − 1 ] (3.604) x = 1 g 1 + ( g t ) 2 − 1 {:(3.604)x=(1)/(g)[sqrt(1+(gt)^(2))-1]:}\begin{equation*} x=\frac{1}{g}\left[\sqrt{1+(g t)^{2}}-1\right] \tag{3.604} \end{equation*}(3.604)x=1g[1+(gt)2−1]
The answer is
(3.605) x = m 0 e E [ 1 + ( e E t m 0 ) 2 − 1 ] (3.605) x = m 0 e E 1 + e E t m 0 2 − 1 {:(3.605)x=(m_(0))/(eE)[sqrt(1+((eEt)/(m_(0)))^(2))-1]:}\begin{equation*} x=\frac{m_{0}}{e E}\left[\sqrt{1+\left(\frac{e E t}{m_{0}}\right)^{2}}-1\right] \tag{3.605} \end{equation*}(3.605)x=m0eE[1+(eEtm0)2−1]
We now want to calculate the velocity as a function of the displacement, we multiply the Lorentz force f = e ( E + u × B ) f = e ( E + u × B ) f=e(E+u xx B)\boldsymbol{f}=e(\boldsymbol{E}+\boldsymbol{u} \times \boldsymbol{B})f=e(E+u×B) by u u u\boldsymbol{u}u and use that the change in kinetic energy is d T = f ⋅ d r d T = f ⋅ d r dT=f*drd T=\boldsymbol{f} \cdot d \boldsymbol{r}dT=f⋅dr. We then introduce the electrostatic potential Φ Φ Phi\PhiΦ ( E = − ∇ Φ ) ( E = − ∇ Φ ) (E=-grad Phi)(\boldsymbol{E}=-\nabla \Phi)(E=−∇Φ) :
(3.606) d T d t = − e ∇ Φ ⋅ u = − e ∇ Φ ⋅ d r d t = − e d Φ d t (3.606) d T d t = − e ∇ Φ ⋅ u = − e ∇ Φ ⋅ d r d t = − e d Φ d t {:(3.606)(dT)/(dt)=-e grad Phi*u=-e grad Phi*(dr)/(dt)=-e(d Phi)/(dt):}\begin{equation*} \frac{d T}{d t}=-e \nabla \Phi \cdot u=-e \nabla \Phi \cdot \frac{d r}{d t}=-e \frac{d \Phi}{d t} \tag{3.606} \end{equation*}(3.606)dTdt=−e∇Φ⋅u=−e∇Φ⋅drdt=−edΦdt
Integration with respect to t t ttt gives
(3.607) T = m 0 ( γ ( u ) − 1 ) = − e Δ Φ = e E r (3.607) T = m 0 ( γ ( u ) − 1 ) = − e Δ Φ = e E r {:(3.607)T=m_(0)(gamma(u)-1)=-e Delta Phi=eEr:}\begin{equation*} T=m_{0}(\gamma(u)-1)=-e \Delta \Phi=e E r \tag{3.607} \end{equation*}(3.607)T=m0(γ(u)−1)=−eΔΦ=eEr
Figure 3.9 A straight uncharged conductor with current I I III as viewed in the inertial systems K K KKK and K ′ K ′ K^(')K^{\prime}K′. Note that K ′ K ′ K^(')K^{\prime}K′ is moving with velocity v v vvv relative to K K KKK along the x 1 x 1 x^(1)x^{1}x1-axis.
i.e.,
(3.608) u ( r ) = 1 − ( 1 + e E r m 0 ) − 2 (3.608) u ( r ) = 1 − 1 + e E r m 0 − 2 {:(3.608)u(r)=sqrt(1-(1+(eEr)/(m_(0)))^(-2)):}\begin{equation*} u(r)=\sqrt{1-\left(1+\frac{e E r}{m_{0}}\right)^{-2}} \tag{3.608} \end{equation*}(3.608)u(r)=1−(1+eErm0)−2

1.119

Introduce K K KKK and K ′ K ′ K^(')K^{\prime}K′ according to Figure 3.9.
a) The components of the electromagnetic field in K K KKK at the point P = ( 0 , 0 , r ) P = ( 0 , 0 , r ) P=(0,0,r)P=(0,0, r)P=(0,0,r) are given by
E i = B 1 = B 3 = 0 (3.609) B 2 = − μ 0 I 2 π r E i = B 1 = B 3 = 0 (3.609) B 2 = − μ 0 I 2 π r {:[E^(i)=B^(1)=B^(3)=0],[(3.609)B^(2)=-(mu_(0)I)/(2pi r)]:}\begin{align*} & E^{i}=B^{1}=B^{3}=0 \\ & B^{2}=-\frac{\mu_{0} I}{2 \pi r} \tag{3.609} \end{align*}Ei=B1=B3=0(3.609)B2=−μ0I2πr
According to the Lorentz transformation formulas for the electromagnetic field
(3.610) E ′ 1 = E 1 , E ′ 2 = γ ( v ) ( E 2 − v B 3 ) , E ′ 3 = γ ( v ) ( E 3 + v B 2 ) B ′ 1 = B 1 , B ′ 2 = γ ( v ) ( B 2 + v c 2 E 3 ) , B ′ 3 = γ ( v ) ( B 3 − v c 2 E 2 ) (3.610) E ′ 1 = E 1 , E ′ 2 = γ ( v ) E 2 − v B 3 , E ′ 3 = γ ( v ) E 3 + v B 2 B ′ 1 = B 1 , B ′ 2 = γ ( v ) B 2 + v c 2 E 3 , B ′ 3 = γ ( v ) B 3 − v c 2 E 2 {:(3.610){:[E^('1)=E^(1)",",E^('2)=gamma(v)(E^(2)-vB^(3))","quadE^('3)=gamma(v)(E^(3)+vB^(2))],[B^('1)=B^(1)",",B^('2)=gamma(v)(B^(2)+(v)/(c^(2))E^(3))","quadB^('3)=gamma(v)(B^(3)-(v)/(c^(2))E^(2))]:}:}\begin{array}{ll} E^{\prime 1}=E^{1}, & E^{\prime 2}=\gamma(v)\left(E^{2}-v B^{3}\right), \quad E^{\prime 3}=\gamma(v)\left(E^{3}+v B^{2}\right) \\ B^{\prime 1}=B^{1}, & B^{\prime 2}=\gamma(v)\left(B^{2}+\frac{v}{c^{2}} E^{3}\right), \quad B^{\prime 3}=\gamma(v)\left(B^{3}-\frac{v}{c^{2}} E^{2}\right) \tag{3.610} \end{array}(3.610)E′1=E1,E′2=γ(v)(E2−vB3),E′3=γ(v)(E3+vB2)B′1=B1,B′2=γ(v)(B2+vc2E3),B′3=γ(v)(B3−vc2E2)
the corresponding components of the electromagnetic field in K ′ K ′ K^(')K^{\prime}K′ are therefore
E ′ 1 = E ′ 2 = 0 , E ′ 3 = − v γ ( v ) μ 0 I 2 π r (3.611) B ′ 1 = B ′ 3 = 0 , B ′ 2 = − γ ( v ) μ 0 I 2 π r E ′ 1 = E ′ 2 = 0 , E ′ 3 = − v γ ( v ) μ 0 I 2 π r (3.611) B ′ 1 = B ′ 3 = 0 , B ′ 2 = − γ ( v ) μ 0 I 2 π r {:[E^('1)=E^('2)=0","quadE^('3)=-(v gamma(v)mu_(0)I)/(2pi r)],[(3.611)B^('1)=B^('3)=0","quadB^('2)=-(gamma(v)mu_(0)I)/(2pi r)]:}\begin{align*} & E^{\prime 1}=E^{\prime 2}=0, \quad E^{\prime 3}=-\frac{v \gamma(v) \mu_{0} I}{2 \pi r} \\ & B^{\prime 1}=B^{\prime 3}=0, \quad B^{\prime 2}=-\frac{\gamma(v) \mu_{0} I}{2 \pi r} \tag{3.611} \end{align*}E′1=E′2=0,E′3=−vγ(v)μ0I2πr(3.611)B′1=B′3=0,B′2=−γ(v)μ0I2πr
b) The current-density 4 -vector in the conductor has the components J μ = J μ = J^(mu)=J^{\mu}=Jμ= ( 0 , c μ 0 I / A , 0 , 0 ) 0 , c μ 0 I / A , 0 , 0 (0,cmu_(0)I//A,0,0)\left(0, c \mu_{0} I / A, 0,0\right)(0,cμ0I/A,0,0) in K K KKK and J ′ μ = ( ρ ′ / ϵ 0 , c μ 0 j ′ 1 , c μ 0 j ′ 2 , c μ 0 j ′ 3 ) J ′ μ = ρ ′ / ϵ 0 , c μ 0 j ′ 1 , c μ 0 j ′ 2 , c μ 0 j ′ 3 J^('mu)=(rho^(')//epsilon_(0),cmu_(0)j^('1),cmu_(0)j^('2),cmu_(0)j^('3))J^{\prime \mu}=\left(\rho^{\prime} / \epsilon_{0}, c \mu_{0} j^{\prime 1}, c \mu_{0} j^{\prime 2}, c \mu_{0} j^{\prime 3}\right)J′μ=(ρ′/ϵ0,cμ0j′1,cμ0j′2,cμ0j′3) in K ′ K ′ K^(')K^{\prime}K′, where J ′ μ = Λ ν μ J v , Λ J ′ μ = Λ ν μ J v , Λ J^('mu)=Lambda_(nu)^(mu)J^(v),LambdaJ^{\prime \mu}=\Lambda_{\nu}^{\mu} J^{v}, \LambdaJ′μ=ΛνμJv,Λ being the Lorentz transformation from K K KKK to K ′ K ′ K^(')K^{\prime}K′ and A A AAA the crosssectional area of the conductor. We then obtain
(3.612) ρ ′ = − v γ ( v ) c 2 I A and j ′ 1 = γ ( v ) I A (3.612) ρ ′ = − v γ ( v ) c 2 I A  and  j ′ 1 = γ ( v ) I A {:(3.612)rho^(')=-(v gamma(v))/(c^(2))(I)/(A)quad" and "quadj^('1)=gamma(v)(I)/(A):}\begin{equation*} \rho^{\prime}=-\frac{v \gamma(v)}{c^{2}} \frac{I}{A} \quad \text { and } \quad j^{\prime 1}=\gamma(v) \frac{I}{A} \tag{3.612} \end{equation*}(3.612)ρ′=−vγ(v)c2IA and j′1=γ(v)IA
Since the cross-sectional area A ′ A ′ A^(')A^{\prime}A′ relative to K ′ K ′ K^(')K^{\prime}K′ is A A AAA, the current relative to K ′ K ′ K^(')K^{\prime}K′ becomes
(3.613) I ′ = A ′ j ′ 1 = γ ( v ) I (3.613) I ′ = A ′ j ′ 1 = γ ( v ) I {:(3.613)I^(')=A^(')j^('1)=gamma(v)I:}\begin{equation*} I^{\prime}=A^{\prime} j^{\prime 1}=\gamma(v) I \tag{3.613} \end{equation*}(3.613)I′=A′j′1=γ(v)I
Now, ρ ′ ρ ′ rho^(')\rho^{\prime}ρ′ and I ′ I ′ I^(')I^{\prime}I′ generate the components of the electromagnetic field in K ′ K ′ K^(')K^{\prime}K′ as
E ′ 1 = E ′ 2 = 0 , E ′ 3 = ρ ′ A ′ 2 π r ′ ϵ 0 = − v γ ( v ) I μ 0 2 π r ′ , (3.614) B ′ 1 = B ′ 3 = 0 , B ′ 2 = − μ 0 I ′ 2 π r ′ E ′ 1 = E ′ 2 = 0 , E ′ 3 = ρ ′ A ′ 2 π r ′ ϵ 0 = − v γ ( v ) I μ 0 2 π r ′ , (3.614) B ′ 1 = B ′ 3 = 0 , B ′ 2 = − μ 0 I ′ 2 π r ′ {:[E^('1)=E^('2)=0","quadE^('3)=(rho^(')A^('))/(2pir^(')epsilon_(0))=-(v gamma(v)Imu_(0))/(2pir^('))","],[(3.614)B^('1)=B^('3)=0","quadB^('2)=-(mu_(0)I^('))/(2pir^('))]:}\begin{align*} & E^{\prime 1}=E^{\prime 2}=0, \quad E^{\prime 3}=\frac{\rho^{\prime} A^{\prime}}{2 \pi r^{\prime} \epsilon_{0}}=-\frac{v \gamma(v) I \mu_{0}}{2 \pi r^{\prime}}, \\ & B^{\prime 1}=B^{\prime 3}=0, \quad B^{\prime 2}=-\frac{\mu_{0} I^{\prime}}{2 \pi r^{\prime}} \tag{3.614} \end{align*}E′1=E′2=0,E′3=ρ′A′2πr′ϵ0=−vγ(v)Iμ02πr′,(3.614)B′1=B′3=0,B′2=−μ0I′2πr′
which is the same result as we obtained in a).

1.120

Inserting the definition of the electromagnetic field strength tensor, i.e., F μ ν = F μ ν = F^(mu nu)=F^{\mu \nu}=Fμν= ∂ μ A ν − ∂ ν A μ ∂ μ A ν − ∂ ν A μ del^(mu)A^(nu)-del^(nu)A^(mu)\partial^{\mu} A^{\nu}-\partial^{\nu} A^{\mu}∂μAν−∂νAμ, into Maxwell's equations ∂ μ F μ ν = J ν ∂ μ F μ ν = J ν del_(mu)F^(mu nu)=J^(nu)\partial_{\mu} F^{\mu \nu}=J^{\nu}∂μFμν=Jν, one obtains
∂ μ F μ v = ∂ μ ( ∂ μ A v − ∂ v A μ ) = { ∂ μ ∂ μ = ◻ and ∂ μ ∂ v = ∂ v ∂ μ } (3.615) = ◻ A v − ∂ v ( ∂ μ A μ ) = J v ∂ μ F μ v = ∂ μ ∂ μ A v − ∂ v A μ = ∂ μ ∂ μ = ◻  and  ∂ μ ∂ v = ∂ v ∂ μ (3.615) = ◻ A v − ∂ v ∂ μ A μ = J v {:[del_(mu)F^(mu v)=del_(mu)(del^(mu)A^(v)-del^(v)A^(mu))={del_(mu)del^(mu)=◻" and "del_(mu)del^(v)=del^(v)del_(mu)}],[(3.615)=◻A^(v)-del^(v)(del_(mu)A^(mu))=J^(v)]:}\begin{align*} \partial_{\mu} F^{\mu v} & =\partial_{\mu}\left(\partial^{\mu} A^{v}-\partial^{v} A^{\mu}\right)=\left\{\partial_{\mu} \partial^{\mu}=\square \text { and } \partial_{\mu} \partial^{v}=\partial^{v} \partial_{\mu}\right\} \\ & =\square A^{v}-\partial^{v}\left(\partial_{\mu} A^{\mu}\right)=J^{v} \tag{3.615} \end{align*}∂μFμv=∂μ(∂μAv−∂vAμ)={∂μ∂μ=◻ and ∂μ∂v=∂v∂μ}(3.615)=◻Av−∂v(∂μAμ)=Jv
Using the Lorenz gauge condition, i.e., ∂ μ A μ = 0 ∂ μ A μ = 0 del_(mu)A^(mu)=0\partial_{\mu} A^{\mu}=0∂μAμ=0, yields
(3.616) ◻ A v = J v (3.616) ◻ A v = J v {:(3.616)◻A^(v)=J^(v):}\begin{equation*} \square A^{v}=J^{v} \tag{3.616} \end{equation*}(3.616)◻Av=Jv
This is the simple form of Maxwell's equations and is a wave equation for A A AAA with source term J J JJJ. Assuming that J = 0 J = 0 J=0J=0J=0, which implies that J v = 0 J v = 0 J^(v)=0J^{v}=0Jv=0, one finds ◻ A v = 0 ◻ A v = 0 ◻A^(v)=0\square A^{v}=0◻Av=0. A useful formula is given by
(3.617) ∂ μ A ν = ∂ μ ( ε ν e i k ⋅ x ) = ε ν ∂ μ e i k ⋅ x = ε ν e i k ⋅ x i k μ = i k μ ε v e i k ⋅ x = i k μ A ν (3.617) ∂ μ A ν = ∂ μ ε ν e i k ⋅ x = ε ν ∂ μ e i k ⋅ x = ε ν e i k ⋅ x i k μ = i k μ ε v e i k ⋅ x = i k μ A ν {:(3.617)del^(mu)A^(nu)=del^(mu)(epsi^(nu)e^(ik*x))=epsi^(nu)del^(mu)e^(ik*x)=epsi^(nu)e^(ik*x)ik^(mu)=ik^(mu)epsi^(v)e^(ik*x)=ik^(mu)A^(nu):}\begin{equation*} \partial^{\mu} A^{\nu}=\partial^{\mu}\left(\varepsilon^{\nu} e^{i k \cdot x}\right)=\varepsilon^{\nu} \partial^{\mu} e^{i k \cdot x}=\varepsilon^{\nu} e^{i k \cdot x} i k^{\mu}=i k^{\mu} \varepsilon^{v} e^{i k \cdot x}=i k^{\mu} A^{\nu} \tag{3.617} \end{equation*}(3.617)∂μAν=∂μ(ενeik⋅x)=εν∂μeik⋅x=ενeik⋅xikμ=ikμεveik⋅x=ikμAν
Using Eq. (3.617), one obtains the electric and magnetic field components as
(3.618) E i = F i 0 = ∂ i A 0 − ∂ 0 A i = i k i ε 0 e i k ⋅ x − i k 0 ε i e i k ⋅ x = i ( k i ε 0 − k 0 ε i ) e i k ⋅ x (3.619) B i = ( ∇ × A ) i = ϵ i j k ∂ j A k = i ϵ i j k k j A k . (3.618) E i = F i 0 = ∂ i A 0 − ∂ 0 A i = i k i ε 0 e i k ⋅ x − i k 0 ε i e i k ⋅ x = i k i ε 0 − k 0 ε i e i k ⋅ x (3.619) B i = ( ∇ × A ) i = ϵ i j k ∂ j A k = i ϵ i j k k j A k . {:[(3.618)E^(i)=F^(i0)=del^(i)A^(0)-del^(0)A^(i)=ik^(i)epsi^(0)e^(ik*x)-ik^(0)epsi^(i)e^(ik*x)=i(k^(i)epsi^(0)-k^(0)epsi^(i))e^(ik*x)],[(3.619)B^(i)=(grad xx A)^(i)=epsilon^(ijk)del^(j)A^(k)=iepsilon^(ijk)k^(j)A^(k).]:}\begin{align*} & E^{i}=F^{i 0}=\partial^{i} A^{0}-\partial^{0} A^{i}=i k^{i} \varepsilon^{0} e^{i k \cdot x}-i k^{0} \varepsilon^{i} e^{i k \cdot x}=i\left(k^{i} \varepsilon^{0}-k^{0} \varepsilon^{i}\right) e^{i k \cdot x} \tag{3.618}\\ & B^{i}=(\nabla \times \boldsymbol{A})^{i}=\epsilon^{i j k} \partial^{j} A^{k}=i \epsilon^{i j k} k^{j} A^{k} . \tag{3.619} \end{align*}(3.618)Ei=Fi0=∂iA0−∂0Ai=ikiε0eik⋅x−ik0εieik⋅x=i(kiε0−k0εi)eik⋅x(3.619)Bi=(∇×A)i=ϵijk∂jAk=iϵijkkjAk.
Multiplying Eqs. (3.618) and (3.619) with k i k i k^(i)k^{i}ki, one finds that
E ⋅ k = E i k i = i ( k i ε 0 − k 0 ε i ) e i k ⋅ x k i = i ( k i k i ε 0 − k 0 k i ε i ) e i k ⋅ x (3.620) = i ( k 2 ε 0 − k 0 k ⋅ ε ) e i k ⋅ x B ⋅ k = B i k i = ϵ i j k i k j A k k i = i ϵ i j k k i k j A k (3.621) = { ϵ i j k is antisymmetric and k i k j is symmetric } = 0 E ⋅ k = E i k i = i k i ε 0 − k 0 ε i e i k ⋅ x k i = i k i k i ε 0 − k 0 k i ε i e i k ⋅ x (3.620) = i k 2 ε 0 − k 0 k ⋅ ε e i k ⋅ x B ⋅ k = B i k i = ϵ i j k i k j A k k i = i ϵ i j k k i k j A k (3.621) = ϵ i j k  is antisymmetric and  k i k j  is symmetric  = 0 {:[E*k=E^(i)k^(i)=i(k^(i)epsi^(0)-k^(0)epsi^(i))e^(ik*x)k^(i)=i(k^(i)k^(i)epsi^(0)-k^(0)k^(i)epsi^(i))e^(ik*x)],[(3.620)=i(k^(2)epsi^(0)-k^(0)k*epsi)e^(ik*x)],[B*k=B^(i)k^(i)=epsilon^(ijk)ik^(j)A^(k)k^(i)=iepsilon^(ijk)k^(i)k^(j)A^(k)],[(3.621)={epsilon^(ijk)" is antisymmetric and "k^(i)k^(j)" is symmetric "}=0]:}\begin{align*} \boldsymbol{E} \cdot \boldsymbol{k} & =E^{i} k^{i}=i\left(k^{i} \varepsilon^{0}-k^{0} \varepsilon^{i}\right) e^{i k \cdot x} k^{i}=i\left(k^{i} k^{i} \varepsilon^{0}-k^{0} k^{i} \varepsilon^{i}\right) e^{i k \cdot x} \\ & =i\left(\boldsymbol{k}^{2} \varepsilon^{0}-k^{0} k \cdot \varepsilon\right) e^{i k \cdot x} \tag{3.620}\\ \boldsymbol{B} \cdot \boldsymbol{k} & =B^{i} k^{i}=\epsilon^{i j k} i k^{j} A^{k} k^{i}=i \epsilon^{i j k} k^{i} k^{j} A^{k} \\ & =\left\{\epsilon^{i j k} \text { is antisymmetric and } k^{i} k^{j} \text { is symmetric }\right\}=0 \tag{3.621} \end{align*}E⋅k=Eiki=i(kiε0−k0εi)eik⋅xki=i(kikiε0−k0kiεi)eik⋅x(3.620)=i(k2ε0−k0k⋅ε)eik⋅xB⋅k=Biki=ϵijkikjAkki=iϵijkkikjAk(3.621)={ϵijk is antisymmetric and kikj is symmetric }=0
Multiplying Eq. (3.617) with η μ ν η μ ν eta_(mu nu)\eta_{\mu \nu}ημν, one has
(3.622) ∂ μ A μ = η μ v ∂ μ A v = η μ v i k μ A v = i k μ A μ = i k μ ε μ e i k ⋅ x = i ( k 0 ε 0 − k ⋅ ε ) e i k ⋅ x (3.622) ∂ μ A μ = η μ v ∂ μ A v = η μ v i k μ A v = i k μ A μ = i k μ ε μ e i k ⋅ x = i k 0 ε 0 − k ⋅ ε e i k ⋅ x {:(3.622)del_(mu)A^(mu)=eta_(mu v)del^(mu)A^(v)=eta_(mu v)ik^(mu)A^(v)=ik_(mu)A^(mu)=ik_(mu)epsi^(mu)e^(ik*x)=i(k_(0)epsi^(0)-k*epsi)e^(ik*x):}\begin{equation*} \partial_{\mu} A^{\mu}=\eta_{\mu v} \partial^{\mu} A^{v}=\eta_{\mu v} i k^{\mu} A^{v}=i k_{\mu} A^{\mu}=i k_{\mu} \varepsilon^{\mu} e^{i k \cdot x}=i\left(k_{0} \varepsilon^{0}-k \cdot \varepsilon\right) e^{i k \cdot x} \tag{3.622} \end{equation*}(3.622)∂μAμ=ημv∂μAv=ημvikμAv=ikμAμ=ikμεμeik⋅x=i(k0ε0−k⋅ε)eik⋅x
but ∂ μ A μ = 0 ∂ μ A μ = 0 del_(mu)A^(mu)=0\partial_{\mu} A^{\mu}=0∂μAμ=0 (Lorenz gauge), so i ( k 0 ε 0 − k ⋅ ε ) e i k ⋅ x = 0 i k 0 ε 0 − k ⋅ ε e i k ⋅ x = 0 i(k_(0)epsi^(0)-k*epsi)e^(ik*x)=0i\left(k_{0} \varepsilon^{0}-k \cdot \varepsilon\right) e^{i k \cdot x}=0i(k0ε0−k⋅ε)eik⋅x=0, i.e.,
(3.623) ε 0 = k ⋅ ε k 0 (3.623) ε 0 = k ⋅ ε k 0 {:(3.623)epsi^(0)=(k*epsi)/(k_(0)):}\begin{equation*} \varepsilon^{0}=\frac{k \cdot \varepsilon}{k_{0}} \tag{3.623} \end{equation*}(3.623)ε0=k⋅εk0
since e i k ⋅ x ≠ 0 e i k ⋅ x ≠ 0 e^(ik*x)!=0e^{i k \cdot x} \neq 0eik⋅x≠0. Inserting Eq. (3.623) into Eq. (3.620) yields
E ⋅ k = i ( k 2 k ⋅ ε k 0 − k 0 k ⋅ ε ) e i k ⋅ x = − i k ⋅ ε k 0 ( k 0 k 0 − k 2 ) e i k ⋅ x = − i k ⋅ ε k 0 k 2 e i k ⋅ x E ⋅ k = i k 2 k ⋅ ε k 0 − k 0 k ⋅ ε e i k ⋅ x = − i k ⋅ ε k 0 k 0 k 0 − k 2 e i k ⋅ x = − i k ⋅ ε k 0 k 2 e i k ⋅ x E*k=i(k^(2)(k*epsi)/(k_(0))-k^(0)k*epsi)e^(ik*x)=-i(k*epsi)/(k_(0))(k_(0)k^(0)-k^(2))e^(ik*x)=-i(k*epsi)/(k_(0))k^(2)e^(ik*x)\boldsymbol{E} \cdot \boldsymbol{k}=i\left(\boldsymbol{k}^{2} \frac{\boldsymbol{k} \cdot \varepsilon}{k_{0}}-k^{0} \boldsymbol{k} \cdot \varepsilon\right) e^{i \boldsymbol{k} \cdot x}=-i \frac{\boldsymbol{k} \cdot \varepsilon}{k_{0}}\left(k_{0} k^{0}-\boldsymbol{k}^{2}\right) e^{i k \cdot x}=-i \frac{\boldsymbol{k} \cdot \varepsilon}{k_{0}} k^{2} e^{i k \cdot x}E⋅k=i(k2k⋅εk0−k0k⋅ε)eik⋅x=−ik⋅εk0(k0k0−k2)eik⋅x=−ik⋅εk0k2eik⋅x.
Taking the partial derivative with respect to x μ x μ x^(mu)x^{\mu}xμ of Eq. (3.617), one finds
◻ A v = ∂ μ ∂ μ A ν = ∂ μ ( i k μ A v ) = i k μ ∂ μ A ν = i k μ ( i k μ A ν ) = − k μ k μ A ν = − k 2 A v ◻ A v = ∂ μ ∂ μ A ν = ∂ μ i k μ A v = i k μ ∂ μ A ν = i k μ i k μ A ν = − k μ k μ A ν = − k 2 A v ◻A^(v)=del_(mu)del^(mu)A^(nu)=del_(mu)(ik^(mu)A^(v))=ik^(mu)del_(mu)A^(nu)=ik^(mu)(ik_(mu)A^(nu))=-k_(mu)k^(mu)A^(nu)=-k^(2)A^(v)\square A^{v}=\partial_{\mu} \partial^{\mu} A^{\nu}=\partial_{\mu}\left(i k^{\mu} A^{v}\right)=i k^{\mu} \partial_{\mu} A^{\nu}=i k^{\mu}\left(i k_{\mu} A^{\nu}\right)=-k_{\mu} k^{\mu} A^{\nu}=-k^{2} A^{v}◻Av=∂μ∂μAν=∂μ(ikμAv)=ikμ∂μAν=ikμ(ikμAν)=−kμkμAν=−k2Av,
but ◻ A v = 0 ◻ A v = 0 ◻A^(v)=0\square A^{v}=0◻Av=0, so − k 2 A v = 0 − k 2 A v = 0 -k^(2)A^(v)=0-k^{2} A^{v}=0−k2Av=0, i.e., k 2 = 0 k 2 = 0 k^(2)=0k^{2}=0k2=0, since A v ≠ 0 A v ≠ 0 A^(v)!=0A^{v} \neq 0Av≠0. Inserting k 2 = 0 k 2 = 0 k^(2)=0k^{2}=0k2=0 into Eq. (3.624) yields
(3.626) E ⋅ k = − i k ⋅ ε k 0 ⋅ 0 ⋅ e i k ⋅ x = 0 (3.626) E ⋅ k = − i k ⋅ ε k 0 ⋅ 0 ⋅ e i k ⋅ x = 0 {:(3.626)E*k=-i(k*epsi)/(k_(0))*0*e^(ik*x)=0:}\begin{equation*} \boldsymbol{E} \cdot \boldsymbol{k}=-i \frac{\boldsymbol{k} \cdot \varepsilon}{k_{0}} \cdot 0 \cdot e^{i k \cdot x}=0 \tag{3.626} \end{equation*}(3.626)E⋅k=−ik⋅εk0⋅0⋅eik⋅x=0
Thus, one obtains
(3.627) E ⋅ k = B ⋅ k = 0 (3.627) E ⋅ k = B ⋅ k = 0 {:(3.627)E*k=B*k=0:}\begin{equation*} \boldsymbol{E} \cdot \boldsymbol{k}=\boldsymbol{B} \cdot \boldsymbol{k}=0 \tag{3.627} \end{equation*}(3.627)E⋅k=B⋅k=0
1.121
Inserting the expression for the free electromagnetic plane wave A μ ( x ) = ε μ e i k ⋅ x A μ ( x ) = ε μ e i k ⋅ x A^(mu)(x)=epsi^(mu)e^(ik*x)A^{\mu}(x)=\varepsilon^{\mu} e^{i k \cdot x}Aμ(x)=εμeik⋅x into the definition of the electromagnetic field strength tensor F μ ν = ∂ μ A ν − ∂ ν A μ F μ ν = ∂ μ A ν − ∂ ν A μ F^(mu nu)=del^(mu)A^(nu)-del^(nu)A^(mu)F^{\mu \nu}=\partial^{\mu} A^{\nu}-\partial^{\nu} A^{\mu}Fμν=∂μAν−∂νAμ, one obtains
(3.628) F μ v = i k μ ε v e i k ⋅ x − i k v ε μ e i k ⋅ x = i ( k μ A v − k ν A μ ) (3.628) F μ v = i k μ ε v e i k ⋅ x − i k v ε μ e i k ⋅ x = i k μ A v − k ν A μ {:(3.628)F^(mu v)=ik^(mu)epsi^(v)e^(ik*x)-ik^(v)epsi^(mu)e^(ik*x)=i(k^(mu)A^(v)-k^(nu)A^(mu)):}\begin{equation*} F^{\mu v}=i k^{\mu} \varepsilon^{v} e^{i k \cdot x}-i k^{v} \varepsilon^{\mu} e^{i k \cdot x}=i\left(k^{\mu} A^{v}-k^{\nu} A^{\mu}\right) \tag{3.628} \end{equation*}(3.628)Fμv=ikμεveik⋅x−ikvεμeik⋅x=i(kμAv−kνAμ)
Thus, we find that
F μ ν F μ ν = − ( k μ A v − k v A μ ) ( k μ A ν − k ν A μ ) = − k 2 A 2 + ( k ⋅ A ) 2 + ( k ⋅ A ) 2 − k 2 A 2 (3.629) = 2 [ ( k ⋅ A ) 2 − k 2 A 2 ] F μ ν F μ ν = − k μ A v − k v A μ k μ A ν − k ν A μ = − k 2 A 2 + ( k ⋅ A ) 2 + ( k ⋅ A ) 2 − k 2 A 2 (3.629) = 2 ( k ⋅ A ) 2 − k 2 A 2 {:[F_(mu nu)F^(mu nu)=-(k_(mu)A_(v)-k_(v)A_(mu))(k^(mu)A^(nu)-k^(nu)A^(mu))],[=-k^(2)A^(2)+(k*A)^(2)+(k*A)^(2)-k^(2)A^(2)],[(3.629)=2[(k*A)^(2)-k^(2)A^(2)]]:}\begin{align*} F_{\mu \nu} F^{\mu \nu} & =-\left(k_{\mu} A_{v}-k_{v} A_{\mu}\right)\left(k^{\mu} A^{\nu}-k^{\nu} A^{\mu}\right) \\ & =-k^{2} A^{2}+(k \cdot A)^{2}+(k \cdot A)^{2}-k^{2} A^{2} \\ & =2\left[(k \cdot A)^{2}-k^{2} A^{2}\right] \tag{3.629} \end{align*}FμνFμν=−(kμAv−kvAμ)(kμAν−kνAμ)=−k2A2+(k⋅A)2+(k⋅A)2−k2A2(3.629)=2[(k⋅A)2−k2A2]
Now, Maxwell's equations ∂ μ F μ ν = J ν ∂ μ F μ ν = J ν del_(mu)F^(mu nu)=J^(nu)\partial_{\mu} F^{\mu \nu}=J^{\nu}∂μFμν=Jν expressed in the 4-vector potential A A AAA when the 4 -current J = 0 J = 0 J=0J=0J=0, i.e., the free case, are ◻ A ν − ∂ v ∂ ⋅ A = 0 ◻ A ν − ∂ v ∂ ⋅ A = 0 ◻A^(nu)-del^(v)del*A=0\square A^{\nu}-\partial^{v} \partial \cdot A=0◻Aν−∂v∂⋅A=0 and in k k kkk-space
(3.630) − k 2 A v + k v k ⋅ A = 0 (3.630) − k 2 A v + k v k ⋅ A = 0 {:(3.630)-k^(2)A^(v)+k^(v)k*A=0:}\begin{equation*} -k^{2} A^{v}+k^{v} k \cdot A=0 \tag{3.630} \end{equation*}(3.630)−k2Av+kvk⋅A=0
Therefore, multiplying the above equation with A ν A ν A_(nu)A_{\nu}Aν, one has
(3.631) ( k ⋅ A ) 2 − k 2 A 2 = 0 (3.631) ( k ⋅ A ) 2 − k 2 A 2 = 0 {:(3.631)(k*A)^(2)-k^(2)A^(2)=0:}\begin{equation*} (k \cdot A)^{2}-k^{2} A^{2}=0 \tag{3.631} \end{equation*}(3.631)(k⋅A)2−k2A2=0
Thus, one obtains
(3.632) F μ ν F μ ν = 2 ⋅ 0 = 0 (3.632) F μ ν F μ ν = 2 ⋅ 0 = 0 {:(3.632)F_(mu nu)F^(mu nu)=2*0=0:}\begin{equation*} F_{\mu \nu} F^{\mu \nu}=2 \cdot 0=0 \tag{3.632} \end{equation*}(3.632)FμνFμν=2⋅0=0
i.e., the invariant F μ ν F μ ν F μ ν F μ ν F_(mu nu)F^(mu nu)F_{\mu \nu} F^{\mu \nu}FμνFμν is zero. Finally, by writing out the invariant ϵ μ ν ω λ F μ ν F ω λ ϵ μ ν ω λ F μ ν F ω λ epsilon_(mu nu omega lambda)F^(mu nu)F^(omega lambda)\epsilon_{\mu \nu \omega \lambda} F^{\mu \nu} F^{\omega \lambda}ϵμνωλFμνFωλ explicitly, one finds that
ϵ μ ν ω λ F μ ν F ω λ = − ϵ μ ν ω λ ( k μ ϵ ν − k ν ϵ μ ) ( k ω ϵ λ − k λ ϵ ω ) e 2 i k ⋅ x = − ϵ μ ν ω λ ( k μ k ω ϵ ν ϵ λ − k μ k λ ϵ ν ϵ ω − k ν k ω ϵ μ ϵ λ + k ν k λ ϵ μ ϵ ω ) e 2 i k ⋅ x (3.633) = 0 ϵ μ ν ω λ F μ ν F ω λ = − ϵ μ ν ω λ k μ ϵ ν − k ν ϵ μ k ω ϵ λ − k λ ϵ ω e 2 i k ⋅ x = − ϵ μ ν ω λ k μ k ω ϵ ν ϵ λ − k μ k λ ϵ ν ϵ ω − k ν k ω ϵ μ ϵ λ + k ν k λ ϵ μ ϵ ω e 2 i k ⋅ x (3.633) = 0 {:[epsilon_(mu nu omega lambda)F^(mu nu)F^(omega lambda)=-epsilon_(mu nu omega lambda)(k^(mu)epsilon^(nu)-k^(nu)epsilon^(mu))(k^(omega)epsilon^(lambda)-k^(lambda)epsilon^(omega))e^(2ik*x)],[=-epsilon_(mu nu omega lambda)(k^(mu)k^(omega)epsilon^(nu)epsilon^(lambda)-k^(mu)k^(lambda)epsilon^(nu)epsilon^(omega)-k^(nu)k^(omega)epsilon^(mu)epsilon^(lambda)+k^(nu)k^(lambda)epsilon^(mu)epsilon^(omega))e^(2ik*x)],[(3.633)=0]:}\begin{align*} \epsilon_{\mu \nu \omega \lambda} F^{\mu \nu} F^{\omega \lambda} & =-\epsilon_{\mu \nu \omega \lambda}\left(k^{\mu} \epsilon^{\nu}-k^{\nu} \epsilon^{\mu}\right)\left(k^{\omega} \epsilon^{\lambda}-k^{\lambda} \epsilon^{\omega}\right) e^{2 i k \cdot x} \\ & =-\epsilon_{\mu \nu \omega \lambda}\left(k^{\mu} k^{\omega} \epsilon^{\nu} \epsilon^{\lambda}-k^{\mu} k^{\lambda} \epsilon^{\nu} \epsilon^{\omega}-k^{\nu} k^{\omega} \epsilon^{\mu} \epsilon^{\lambda}+k^{\nu} k^{\lambda} \epsilon^{\mu} \epsilon^{\omega}\right) e^{2 i k \cdot x} \\ & =0 \tag{3.633} \end{align*}ϵμνωλFμνFωλ=−ϵμνωλ(kμϵν−kνϵμ)(kωϵλ−kλϵω)e2ik⋅x=−ϵμνωλ(kμkωϵνϵλ−kμkλϵνϵω−kνkωϵμϵλ+kνkλϵμϵω)e2ik⋅x(3.633)=0
since ϵ μ ν ω λ ϵ μ ν ω λ epsilon_(mu nu omega lambda)\epsilon_{\mu \nu \omega \lambda}ϵμνωλ is totally antisymmetric and each term inside the parenthesis has two pairs of symmetric indices, i.e., the invariant ϵ μ ν ω λ F μ ν F ω λ ϵ μ ν ω λ F μ ν F ω λ epsilon_(mu nu omega lambda)F^(mu nu)F^(omega lambda)\epsilon_{\mu \nu \omega \lambda} F^{\mu \nu} F^{\omega \lambda}ϵμνωλFμνFωλ is also zero.
The result shows that for a plane wave solution of Maxwell's equations, the electric and magnetic fields oscillate in such a way that their magnitude is always equal (if multiplying the magnetic field with c 2 c 2 c^(2)c^{2}c2 to get the correct units, the invariant F μ ν F μ ν F μ ν F μ ν F_(mu nu)F^(mu nu)F_{\mu \nu} F^{\mu \nu}FμνFμν is proportional to E 2 − c 2 B 2 E 2 − c 2 B 2 E^(2)-c^(2)B^(2)\boldsymbol{E}^{2}-c^{2} \boldsymbol{B}^{2}E2−c2B2 ) and that they are always orthogonal ( ϵ μ ν ω λ F μ ν F ω λ ϵ μ ν ω λ F μ ν F ω λ (epsilon_(mu nu omega lambda)F^(mu nu)F^(omega lambda):}\left(\epsilon_{\mu \nu \omega \lambda} F^{\mu \nu} F^{\omega \lambda}\right.(ϵμνωλFμνFωλ is proportional to E ⋅ B ) E ⋅ B {:E*B)\left.\boldsymbol{E} \cdot \boldsymbol{B}\right)E⋅B).

1.122

a) Using the Lorentz invariant ϵ μ ν λ ω F μ ν F λ ω = − 8 c E ⋅ B ϵ μ ν λ ω F μ ν F λ ω = − 8 c E ⋅ B epsilon_(mu nu lambda omega)F^(mu nu)F^(lambda omega)=-8cE*B\epsilon_{\mu \nu \lambda \omega} F^{\mu \nu} F^{\lambda \omega}=-8 c \boldsymbol{E} \cdot \boldsymbol{B}ϵμνλωFμνFλω=−8cE⋅B, one finds
(3.634) E ⋅ B = − 1 8 c ϵ μ ν λ ω F μ ν F λ ω (3.634) E ⋅ B = − 1 8 c ϵ μ ν λ ω F μ ν F λ ω {:(3.634)E*B=-(1)/(8c)epsilon_(mu nu lambda omega)F^(mu nu)F^(lambda omega):}\begin{equation*} \boldsymbol{E} \cdot \boldsymbol{B}=-\frac{1}{8 c} \epsilon_{\mu \nu \lambda \omega} F^{\mu \nu} F^{\lambda \omega} \tag{3.634} \end{equation*}(3.634)E⋅B=−18cϵμνλωFμνFλω
where ϵ μ ν λ ω ϵ μ ν λ ω epsilon^(mu nu lambda omega)\epsilon^{\mu \nu \lambda \omega}ϵμνλω is a totally antisymmetric fourth rank tensor with ϵ 0123 = 1 ϵ 0123 = 1 epsilon^(0123)=1\epsilon^{0123}=1ϵ0123=1. For any vector index μ μ mu\muμ, the inner product A μ B μ A μ B μ A^(mu)B_(mu)A^{\mu} B_{\mu}AμBμ is Lorentz invariant, and for a Lorentz transformation Λ Λ Lambda\LambdaΛ with det Λ = 1 det Λ = 1 det Lambda=1\operatorname{det} \Lambda=1detΛ=1, it holds that ϵ ′ μ ν λ ω = ϵ μ ν λ ω ϵ ′ μ ν λ ω = ϵ μ ν λ ω epsilon^('mu nu lambda omega)=epsilon^(mu nu lambda omega)\epsilon^{\prime \mu \nu \lambda \omega}=\epsilon^{\mu \nu \lambda \omega}ϵ′μνλω=ϵμνλω. Thus, one has
(3.635) ϵ μ ν λ ω ′ F ′ μ ν F ′ λ ω = ϵ μ ν λ ω F ′ μ ν F ′ λ ω = ϵ μ ν λ ω F μ ν F λ ω (3.635) ϵ μ ν λ ω ′ F ′ μ ν F ′ λ ω = ϵ μ ν λ ω F ′ μ ν F ′ λ ω = ϵ μ ν λ ω F μ ν F λ ω {:(3.635)epsilon_(mu nu lambda omega)^(')F^('mu nu)F^('lambda omega)=epsilon_(mu nu lambda omega)F^('mu nu)F^('lambda omega)=epsilon_(mu nu lambda omega)F^(mu nu)F^(lambda omega):}\begin{equation*} \epsilon_{\mu \nu \lambda \omega}^{\prime} F^{\prime \mu \nu} F^{\prime \lambda \omega}=\epsilon_{\mu \nu \lambda \omega} F^{\prime \mu \nu} F^{\prime \lambda \omega}=\epsilon_{\mu \nu \lambda \omega} F^{\mu \nu} F^{\lambda \omega} \tag{3.635} \end{equation*}(3.635)ϵμνλω′F′μνF′λω=ϵμνλωF′μνF′λω=ϵμνλωFμνFλω
Finally, using Eq. (3.634), one obtains
(3.636) E ′ ⋅ B ′ = E ⋅ B (3.636) E ′ ⋅ B ′ = E ⋅ B {:(3.636)E^(')*B^(')=E*B:}\begin{equation*} \boldsymbol{E}^{\prime} \cdot B^{\prime}=E \cdot B \tag{3.636} \end{equation*}(3.636)E′⋅B′=E⋅B
b) The expression ϵ α β γ δ F α β F γ δ ϵ α β γ δ F α β F γ δ epsilon_(alpha beta gamma delta)F^(alpha beta)F^(gamma delta)\epsilon_{\alpha \beta \gamma \delta} F^{\alpha \beta} F^{\gamma \delta}ϵαβγδFαβFγδ is Lorentz invariant and equal to − 8 c E ⋅ B − 8 c E ⋅ B -8cE*B-8 c \boldsymbol{E} \cdot \boldsymbol{B}−8cE⋅B. Thus, if E ⋅ B = 0 E ⋅ B = 0 E*B=0\boldsymbol{E} \cdot \boldsymbol{B}=0E⋅B=0 for one observer, then it is zero for any observer in inertial frames.
c) From Problem 1.120 we have E i = i ( k i ε 0 − k 0 ε i ) e i k ⋅ x E i = i k i ε 0 − k 0 ε i e i k ⋅ x E^(i)=i(k^(i)epsi^(0)-k^(0)epsi^(i))e^(ik*x)E^{i}=i\left(k^{i} \varepsilon^{0}-k^{0} \varepsilon^{i}\right) e^{i k \cdot x}Ei=i(kiε0−k0εi)eik⋅x and B i = i ϵ i j k k j ε k e i k ⋅ x B i = i ϵ i j k k j ε k e i k ⋅ x B^(i)=iepsilon^(ijk)k^(j)epsi^(k)e^(ik*x)B^{i}=i \epsilon^{i j k} k^{j} \varepsilon^{k} e^{i k \cdot x}Bi=iϵijkkjεkeik⋅x. This gives immediately that E ⋅ B = E i B i = 0 E ⋅ B = E i B i = 0 E*B=E^(i)B^(i)=0\boldsymbol{E} \cdot \boldsymbol{B}=E^{i} B^{i}=0E⋅B=EiBi=0, since ϵ i j k ϵ i j k epsilon^(ijk)\epsilon^{i j k}ϵijk is antisymmetric.
d) The components of E × B E × B E xx B\boldsymbol{E} \times \boldsymbol{B}E×B can be written as ( E × B ) i = e 2 i k ⋅ x ( k i M + ε i N ) ( E × B ) i = e 2 i k ⋅ x k i M + ε i N (E xx B)^(i)=e^(2ik*x)(k^(i)M+epsi^(i)N)(\boldsymbol{E} \times \boldsymbol{B})^{i}=e^{2 i k \cdot x}\left(k^{i} M+\varepsilon^{i} N\right)(E×B)i=e2ik⋅x(kiM+εiN) with M = k 0 ε 2 − k ⋅ ε ε 0 M = k 0 ε 2 − k ⋅ ε ε 0 M=k^(0)epsi^(2)-k*epsiepsi^(0)M=k^{0} \varepsilon^{2}-\boldsymbol{k} \cdot \varepsilon \varepsilon^{0}M=k0ε2−k⋅εε0 and N = k 2 ε 0 − k 0 k ⋅ ε N = k 2 ε 0 − k 0 k ⋅ ε N=k^(2)epsi^(0)-k^(0)k*epsiN=\boldsymbol{k}^{2} \varepsilon^{0}-k^{0} \boldsymbol{k} \cdot \varepsilonN=k2ε0−k0k⋅ε. Choose k = ( k 0 , k 0 , 0 , 0 ) k = k 0 , k 0 , 0 , 0 k=(k^(0),k^(0),0,0)k=\left(k^{0}, k^{0}, 0,0\right)k=(k0,k0,0,0) and ε = ( 0 , 0 , 1 , 0 ) ε = ( 0 , 0 , 1 , 0 ) epsi=(0,0,1,0)\varepsilon=(0,0,1,0)ε=(0,0,1,0) or ε = ( 0 , 0 , 0 , 1 ) ε = ( 0 , 0 , 0 , 1 ) epsi=(0,0,0,1)\varepsilon=(0,0,0,1)ε=(0,0,0,1). Then, it holds that E × B = k 0 ( k 0 , 0 , 0 ) e 2 i k ⋅ x E × B = k 0 k 0 , 0 , 0 e 2 i k ⋅ x E xx B=k^(0)(k^(0),0,0)e^(2ik*x)\boldsymbol{E} \times \boldsymbol{B}=k^{0}\left(k^{0}, 0,0\right) e^{2 i k \cdot x}E×B=k0(k0,0,0)e2ik⋅x,
showing that only the 1-component of E × B E × B E xx B\boldsymbol{E} \times \boldsymbol{B}E×B is nonvanishing, and thus, the product is proportional to k = ( k 0 , 0 , 0 ) k = k 0 , 0 , 0 k=(k^(0),0,0)\boldsymbol{k}=\left(k^{0}, 0,0\right)k=(k0,0,0).

1.123

Using the Lorentz force law, we have
(3.637) d p d t = − e ( u × B ) (3.637) d p d t = − e ( u × B ) {:(3.637)(dp)/(dt)=-e(u xx B):}\begin{equation*} \frac{d p}{d t}=-e(\boldsymbol{u} \times \boldsymbol{B}) \tag{3.637} \end{equation*}(3.637)dpdt=−e(u×B)
where e e eee is the electron charge and p = m 0 c x ˙ = m 0 u γ p = m 0 c x ˙ = m 0 u γ p=m_(0)cx^(˙)=m_(0)u gamma\boldsymbol{p}=m_{0} c \dot{\boldsymbol{x}}=m_{0} \boldsymbol{u} \gammap=m0cx˙=m0uγ with γ ≡ 1 / 1 − u 2 / c 2 γ ≡ 1 / 1 − u 2 / c 2 gamma-=1//sqrt(1-u^(2)//c^(2))\gamma \equiv 1 / \sqrt{1-u^{2} / c^{2}}γ≡1/1−u2/c2. From the Lorentz force law, we deduce that
(3.638) d p d t ⋅ p = 0 (3.638) d p d t ⋅ p = 0 {:(3.638)(dp)/(dt)*p=0:}\begin{equation*} \frac{d \boldsymbol{p}}{d t} \cdot \boldsymbol{p}=0 \tag{3.638} \end{equation*}(3.638)dpdt⋅p=0
which implies that p 2 = m 0 2 u 2 γ 2 = p 2 = m 0 2 u 2 γ 2 = p^(2)=m_(0)^(2)u^(2)gamma^(2)=\boldsymbol{p}^{2}=m_{0}^{2} u^{2} \gamma^{2}=p2=m02u2γ2= const. Since | u | | u | |u||\boldsymbol{u}||u| is constant and the magnetic field B = ( 0 , 0 , B ) B = ( 0 , 0 , B ) B=(0,0,B)\boldsymbol{B}=(0,0, B)B=(0,0,B) is symmetric around the z z zzz-axis, we make the ansatz
(3.639) u ( t ) = ( u cos ⁡ ( α t ) , u sin ⁡ ( α t ) , 0 ) (3.639) u ( t ) = ( u cos ⁡ ( α t ) , u sin ⁡ ( α t ) , 0 ) {:(3.639)u(t)=(u cos(alpha t)","u sin(alpha t)","0):}\begin{equation*} \boldsymbol{u}(t)=(u \cos (\alpha t), u \sin (\alpha t), 0) \tag{3.639} \end{equation*}(3.639)u(t)=(ucos⁡(αt),usin⁡(αt),0)
which is automatically consistent with the initial condition u ( t = 0 ) = ( u , 0 , 0 ) u ( t = 0 ) = ( u , 0 , 0 ) u(t=0)=(u,0,0)\boldsymbol{u}(t=0)=(u, 0,0)u(t=0)=(u,0,0). Now, the Lorentz force law for the components is
(3.640) d p x d t = − e B u y , d p y d t = e B u x , d p z d t = 0 (3.640) d p x d t = − e B u y , d p y d t = e B u x , d p z d t = 0 {:(3.640)(dp_(x))/(dt)=-eBu_(y)","quad(dp_(y))/(dt)=eBu_(x)","quad(dp_(z))/(dt)=0:}\begin{equation*} \frac{d p_{x}}{d t}=-e B u_{y}, \quad \frac{d p_{y}}{d t}=e B u_{x}, \quad \frac{d p_{z}}{d t}=0 \tag{3.640} \end{equation*}(3.640)dpxdt=−eBuy,dpydt=eBux,dpzdt=0
which after inserting the ansatz yields
(3.641) d p x d t = − e B u sin ⁡ ( α t ) , d p y d t = e B u cos ⁡ ( α t ) , d p z d t = 0 (3.641) d p x d t = − e B u sin ⁡ ( α t ) , d p y d t = e B u cos ⁡ ( α t ) , d p z d t = 0 {:(3.641)(dp_(x))/(dt)=-eBu sin(alpha t)","quad(dp_(y))/(dt)=eBu cos(alpha t)","quad(dp_(z))/(dt)=0:}\begin{equation*} \frac{d p_{x}}{d t}=-e B u \sin (\alpha t), \quad \frac{d p_{y}}{d t}=e B u \cos (\alpha t), \quad \frac{d p_{z}}{d t}=0 \tag{3.641} \end{equation*}(3.641)dpxdt=−eBusin⁡(αt),dpydt=eBucos⁡(αt),dpzdt=0
Then, integrating these components leads to
(3.642) p x = e B u α ⋅ cos ⁡ ( α t ) + c 1 , p y = e B u α sin ⁡ ( α t ) + c 2 , p z = c 3 , (3.642) p x = e B u α ⋅ cos ⁡ ( α t ) + c 1 , p y = e B u α sin ⁡ ( α t ) + c 2 , p z = c 3 , {:(3.642)p_(x)=(eBu)/(alpha)*cos(alpha t)+c_(1)","quadp_(y)=(eBu)/(alpha)sin(alpha t)+c_(2)","quadp_(z)=c_(3)",":}\begin{equation*} p_{x}=\frac{e B u}{\alpha} \cdot \cos (\alpha t)+c_{1}, \quad p_{y}=\frac{e B u}{\alpha} \sin (\alpha t)+c_{2}, \quad p_{z}=c_{3}, \tag{3.642} \end{equation*}(3.642)px=eBuα⋅cos⁡(αt)+c1,py=eBuαsin⁡(αt)+c2,pz=c3,
where c 1 , c 2 c 1 , c 2 c_(1),c_(2)c_{1}, c_{2}c1,c2, and c 3 c 3 c_(3)c_{3}c3 are integration constants. Consistency with p = m 0 u γ p = m 0 u γ p=m_(0)u gamma\boldsymbol{p}=m_{0} \boldsymbol{u} \gammap=m0uγ means that c 1 = c 2 = c 3 = 0 c 1 = c 2 = c 3 = 0 c_(1)=c_(2)=c_(3)=0c_{1}=c_{2}=c_{3}=0c1=c2=c3=0 and α = e B / ( m 0 γ ) α = e B / m 0 γ alpha=eB//(m_(0)gamma)\alpha=e B /\left(m_{0} \gamma\right)α=eB/(m0γ). Thus, integrating the expression for u u u\boldsymbol{u}u, we obtain the trajectory of the electron as
(3.643) x ( t ) = u m 0 γ e B ( sin ⁡ e B t m 0 γ , − cos ⁡ e B t m 0 γ , 0 ) + x 0 (3.643) x ( t ) = u m 0 γ e B sin ⁡ e B t m 0 γ , − cos ⁡ e B t m 0 γ , 0 + x 0 {:(3.643)x(t)=(um_(0)gamma)/(eB)(sin((eBt)/(m_(0)gamma)),-cos((eBt)/(m_(0)gamma)),0)+x_(0):}\begin{equation*} \boldsymbol{x}(t)=\frac{u m_{0} \gamma}{e B}\left(\sin \frac{e B t}{m_{0} \gamma},-\cos \frac{e B t}{m_{0} \gamma}, 0\right)+x_{0} \tag{3.643} \end{equation*}(3.643)x(t)=um0γeB(sin⁡eBtm0γ,−cos⁡eBtm0γ,0)+x0
where x 0 x 0 x_(0)x_{0}x0 is an integration constant. This trajectory is the equation for a circle perpendicular to the z z zzz-axis with center at x 0 x 0 x_(0)\boldsymbol{x}_{0}x0 and radius r = | x − x 0 | = u m 0 γ / ( e B ) r = x − x 0 = u m 0 γ / ( e B ) r=|x-x_(0)|=um_(0)gamma//(eB)r=\left|\boldsymbol{x}-\boldsymbol{x}_{0}\right|=u m_{0} \gamma /(e B)r=|x−x0|=um0γ/(eB). The time for one revolution is t 0 = 2 π m 0 γ / ( e B ) t 0 = 2 π m 0 γ / ( e B ) t_(0)=2pim_(0)gamma//(eB)t_{0}=2 \pi m_{0} \gamma /(e B)t0=2πm0γ/(eB).

1.124

The two quantities E ⋅ B E ⋅ B E*B\mathbf{E} \cdot \mathbf{B}E⋅B and E 2 − c 2 B 2 E 2 − c 2 B 2 E^(2)-c^(2)B^(2)\mathbf{E}^{2}-c^{2} \mathbf{B}^{2}E2−c2B2 are Lorentz invariants, where E ⋅ B = 0 E ⋅ B = 0 E*B=0\mathbf{E} \cdot \mathbf{B}=0E⋅B=0 and E 2 − c 2 B 2 = 0 E 2 − c 2 B 2 = 0 E^(2)-c^(2)B^(2)=0\mathbf{E}^{2}-c^{2} \mathbf{B}^{2}=0E2−c2B2=0, since E = ( c B , 0 , 0 ) E = ( c B , 0 , 0 ) E=(cB,0,0)\mathbf{E}=(c B, 0,0)E=(cB,0,0) and B = ( 0 , B , 0 ) B = ( 0 , B , 0 ) B=(0,B,0)\mathbf{B}=(0, B, 0)B=(0,B,0). Therefore, one has
E ′ ⋅ B ′ = 0 E ′ ⋅ B ′ = 0 E^(')*B^(')=0\mathbf{E}^{\prime} \cdot \mathbf{B}^{\prime}=0E′⋅B′=0 and E ′ 2 − c 2 B ′ 2 = 0 E ′ 2 − c 2 B ′ 2 = 0 E^('2)-c^(2)B^('2)=0\mathbf{E}^{\prime 2}-c^{2} \mathbf{B}^{\prime 2}=0E′2−c2B′2=0. Inserting E ′ = ( 0 , 2 c B , c B ) E ′ = ( 0 , 2 c B , c B ) E^(')=(0,2cB,cB)\mathbf{E}^{\prime}=(0,2 c B, c B)E′=(0,2cB,cB) and B ′ = ( 0 , B y ′ , B z ′ ) B ′ = 0 , B y ′ , B z ′ B^(')=(0,B_(y)^('),B_(z)^('))\mathbf{B}^{\prime}=\left(0, B_{y}^{\prime}, B_{z}^{\prime}\right)B′=(0,By′,Bz′), one obtains
(3.644) { B y ′ 2 + B z ′ 2 = 5 B 2 2 B y ′ + B z ′ = 0 (3.644) B y ′ 2 + B z ′ 2 = 5 B 2 2 B y ′ + B z ′ = 0 {:(3.644){[B_(y)^('2)+B_(z)^('2)=5B^(2)],[2B_(y)^(')+B_(z)^(')=0]:}:}\left\{\begin{array}{l} B_{y}^{\prime 2}+B_{z}^{\prime 2}=5 B^{2} \tag{3.644}\\ 2 B_{y}^{\prime}+B_{z}^{\prime}=0 \end{array}\right.(3.644){By′2+Bz′2=5B22By′+Bz′=0
Solving this system of equations, one finds that B y ′ = ± B B y ′ = ± B B_(y)^(')=+-BB_{y}^{\prime}= \pm BBy′=±B and B z ′ = ∓ 2 B B z ′ = ∓ 2 B B_(z)^(')=∓2BB_{z}^{\prime}=\mp 2 BBz′=∓2B. Thus, the answer is B ′ = ± ( 0 , B , − 2 B ) B ′ = ± ( 0 , B , − 2 B ) B^(')=+-(0,B,-2B)\mathbf{B}^{\prime}= \pm(0, B,-2 B)B′=±(0,B,−2B).
1.125
The two quantities E 2 − c 2 B 2 E 2 − c 2 B 2 E^(2)-c^(2)B^(2)\mathbf{E}^{2}-c^{2} \mathbf{B}^{2}E2−c2B2 and E ⋅ B E ⋅ B E*B\mathbf{E} \cdot \mathbf{B}E⋅B are Lorentz invariants, where E 2 − c 2 B 2 = E 2 − c 2 B 2 = E^(2)-c^(2)B^(2)=\mathbf{E}^{2}-c^{2} \mathbf{B}^{2}=E2−c2B2= − 2 α 2 − 2 α 2 -2alpha^(2)-2 \alpha^{2}−2α2 and E ⋅ B = 0 E ⋅ B = 0 E*B=0\mathbf{E} \cdot \mathbf{B}=0E⋅B=0, since E = ( α , − α , 0 ) E = ( α , − α , 0 ) E=(alpha,-alpha,0)\mathbf{E}=(\alpha,-\alpha, 0)E=(α,−α,0) and B = ( 0 , 0 , 2 α / c ) B = ( 0 , 0 , 2 α / c ) B=(0,0,2alpha//c)\mathbf{B}=(0,0,2 \alpha / c)B=(0,0,2α/c), where α ≠ 0 α ≠ 0 alpha!=0\alpha \neq 0α≠0. Therefore, one has E ′ 2 − c 2 B ′ 2 = − 2 α 2 E ′ 2 − c 2 B ′ 2 = − 2 α 2 E^('2)-c^(2)B^('2)=-2alpha^(2)\mathbf{E}^{\prime 2}-c^{2} \mathbf{B}^{\prime 2}=-2 \alpha^{2}E′2−c2B′2=−2α2 and E ′ ⋅ B ′ = 0 E ′ ⋅ B ′ = 0 E^(')*B^(')=0\mathbf{E}^{\prime} \cdot \mathbf{B}^{\prime}=0E′⋅B′=0. Inserting E ′ = ( 0 , 0 , 2 α ) E ′ = ( 0 , 0 , 2 α ) E^(')=(0,0,2alpha)\mathbf{E}^{\prime}=(0,0,2 \alpha)E′=(0,0,2α) and B ′ = ( B x ′ , α / c , B z ′ ) B ′ = B x ′ , α / c , B z ′ B^(')=(B_(x)^('),alpha//c,B_(z)^('))\mathbf{B}^{\prime}=\left(B_{x}^{\prime}, \alpha / c, B_{z}^{\prime}\right)B′=(Bx′,α/c,Bz′), one obtains
(3.645) { B x ′ 2 + B z ′ 2 = 5 α 2 c 2 2 α B z ′ = 0 (3.645) B x ′ 2 + B z ′ 2 = 5 α 2 c 2 2 α B z ′ = 0 {:(3.645){[B_(x)^('2)+B_(z)^('2)=(5alpha^(2))/(c^(2))],[2alphaB_(z)^(')=0]:}:}\left\{\begin{array}{l} B_{x}^{\prime 2}+B_{z}^{\prime 2}=\frac{5 \alpha^{2}}{c^{2}} \tag{3.645}\\ 2 \alpha B_{z}^{\prime}=0 \end{array}\right.(3.645){Bx′2+Bz′2=5α2c22αBz′=0
Solving this system of equations, one finds that B x ′ = ± α 5 c B x ′ = ± α 5 c B_(x)^(')=+-(alphasqrt5)/(c)B_{x}^{\prime}= \pm \frac{\alpha \sqrt{5}}{c}Bx′=±α5c and B z ′ = 0 B z ′ = 0 B_(z)^(')=0B_{z}^{\prime}=0Bz′=0. Thus, it follows that B ′ = ( ± α 5 / c , α / c , 0 ) B ′ = ( ± α 5 / c , α / c , 0 ) B^(')=(+-alphasqrt5//c,alpha//c,0)\mathbf{B}^{\prime}=( \pm \alpha \sqrt{5} / c, \alpha / c, 0)B′=(±α5/c,α/c,0).

1.126

The two quantities E 2 − c 2 B 2 E 2 − c 2 B 2 E^(2)-c^(2)B^(2)\mathbf{E}^{2}-c^{2} \mathbf{B}^{2}E2−c2B2 and E ⋅ B E ⋅ B E*B\mathbf{E} \cdot \mathbf{B}E⋅B are Lorentz invariants, where E ⋅ B = 0 E ⋅ B = 0 E*B=0\mathbf{E} \cdot \mathbf{B}=0E⋅B=0 and E 2 − c 2 B 2 = − 2 β 2 E 2 − c 2 B 2 = − 2 β 2 E^(2)-c^(2)B^(2)=-2beta^(2)\mathbf{E}^{2}-c^{2} \mathbf{B}^{2}=-2 \beta^{2}E2−c2B2=−2β2. Therefore, one has E ′ − c 2 B ′ = − 2 β 2 E ′ − c 2 B ′ = − 2 β 2 E^(')-c^(2)B^(')=-2beta^(2)\mathbf{E}^{\prime}-c^{2} \mathbf{B}^{\prime}=-2 \beta^{2}E′−c2B′=−2β2 and E ′ ⋅ B ′ = 0 E ′ ⋅ B ′ = 0 E^(')*B^(')=0\mathbf{E}^{\prime} \cdot \mathbf{B}^{\prime}=0E′⋅B′=0. Inserting E ′ = ( 2 β , 0 , 0 ) E ′ = ( 2 β , 0 , 0 ) E^(')=(2beta,0,0)\mathbf{E}^{\prime}=(2 \beta, 0,0)E′=(2β,0,0) and B ′ = ( B x ′ , B y ′ , β / c ) B ′ = B x ′ , B y ′ , β / c B^(')=(B_(x)^('),B_(y)^('),beta//c)\mathbf{B}^{\prime}=\left(B_{x}^{\prime}, B_{y}^{\prime}, \beta / c\right)B′=(Bx′,By′,β/c), one obtains
(3.646) { c 2 ( B x ′ 2 + B y ′ 2 ) = − 5 β 2 2 β B x ′ = 0 (3.646) c 2 B x ′ 2 + B y ′ 2 = − 5 β 2 2 β B x ′ = 0 {:(3.646){[c^(2)(B_(x)^('2)+B_(y)^('2))=-5beta^(2)],[2betaB_(x)^(')=0]:}:}\left\{\begin{array}{l} c^{2}\left(B_{x}^{\prime 2}+B_{y}^{\prime 2}\right)=-5 \beta^{2} \tag{3.646}\\ 2 \beta B_{x}^{\prime}=0 \end{array}\right.(3.646){c2(Bx′2+By′2)=−5β22βBx′=0
Solving this system of equations, one finds that B x ′ = 0 B x ′ = 0 B_(x)^(')=0B_{x}^{\prime}=0Bx′=0 and B y ′ = ± 5 β / c B y ′ = ± 5 β / c B_(y)^(')=+-sqrt5beta//cB_{y}^{\prime}= \pm \sqrt{5} \beta / cBy′=±5β/c. Thus, it follows that B ′ = ( 0 , ± 5 β / c , β / c ) B ′ = ( 0 , ± 5 β / c , β / c ) B^(')=(0,+-sqrt5beta//c,beta//c)\mathbf{B}^{\prime}=(0, \pm \sqrt{5} \beta / c, \beta / c)B′=(0,±5β/c,β/c).

1.127

Observer A A AAA measures the electric and magnetic fields to be E = ( α , 0 , 0 ) E = ( α , 0 , 0 ) E=(alpha,0,0)\mathbf{E}=(\alpha, 0,0)E=(α,0,0) and B = B = B=\mathbf{B}=B= ( α / c , 0 , 2 α / c ) ( α / c , 0 , 2 α / c ) (alpha//c,0,2alpha//c)(\alpha / c, 0,2 \alpha / c)(α/c,0,2α/c), respectively, where α ≠ 0 α ≠ 0 alpha!=0\alpha \neq 0α≠0. The two quantities E 2 − c 2 B 2 E 2 − c 2 B 2 E^(2)-c^(2)B^(2)\mathbf{E}^{2}-c^{2} \mathbf{B}^{2}E2−c2B2 and E ⋅ B E ⋅ B E*B\mathbf{E} \cdot \mathbf{B}E⋅B are Lorentz invariants, where E 2 − c 2 B 2 = − 4 α 2 E 2 − c 2 B 2 = − 4 α 2 E^(2)-c^(2)B^(2)=-4alpha^(2)\mathbf{E}^{2}-c^{2} \mathbf{B}^{2}=-4 \alpha^{2}E2−c2B2=−4α2 and E ⋅ B = α 2 / c E ⋅ B = α 2 / c E*B=alpha^(2)//c\mathbf{E} \cdot \mathbf{B}=\alpha^{2} / cE⋅B=α2/c. Therefore, one has E ′ 2 − c 2 B ′ 2 = − 4 α 2 E ′ 2 − c 2 B ′ 2 = − 4 α 2 E^('2)-c^(2)B^('2)=-4alpha^(2)\mathbf{E}^{\prime 2}-c^{2} \mathbf{B}^{\prime 2}=-4 \alpha^{2}E′2−c2B′2=−4α2 and E ′ ⋅ B ′ = α 2 / c E ′ ⋅ B ′ = α 2 / c E^(')*B^(')=alpha^(2)//c\mathbf{E}^{\prime} \cdot \mathbf{B}^{\prime}=\alpha^{2} / cE′⋅B′=α2/c. Inserting E ′ = ( E x ′ , α , 0 ) E ′ = E x ′ , α , 0 E^(')=(E_(x)^('),alpha,0)\mathbf{E}^{\prime}=\left(E_{x}^{\prime}, \alpha, 0\right)E′=(Ex′,α,0) and B ′ = ( α / c , B y ′ , α / c ) B ′ = α / c , B y ′ , α / c B^(')=(alpha//c,B_(y)^('),alpha//c)\mathbf{B}^{\prime}=\left(\alpha / c, B_{y}^{\prime}, \alpha / c\right)B′=(α/c,By′,α/c), one obtains
(3.647) { E x ′ 2 − c 2 B y ′ 2 = − 3 α 2 E x ′ + c B y ′ = α (3.647) E x ′ 2 − c 2 B y ′ 2 = − 3 α 2 E x ′ + c B y ′ = α {:(3.647){[E_(x)^('2)-c^(2)B_(y)^('2)=-3alpha^(2)],[E_(x)^(')+cB_(y)^(')=alpha]:}:}\left\{\begin{array}{l} E_{x}^{\prime 2}-c^{2} B_{y}^{\prime 2}=-3 \alpha^{2} \tag{3.647}\\ E_{x}^{\prime}+c B_{y}^{\prime}=\alpha \end{array}\right.(3.647){Ex′2−c2By′2=−3α2Ex′+cBy′=α
Solving this system of equations, one finds that E x ′ = − α E x ′ = − α E_(x)^(')=-alphaE_{x}^{\prime}=-\alphaEx′=−α and B y ′ = 2 α / c B y ′ = 2 α / c B_(y)^(')=2alpha//cB_{y}^{\prime}=2 \alpha / cBy′=2α/c. Thus, it holds that E ′ = ( − α , α , 0 ) E ′ = ( − α , α , 0 ) E^(')=(-alpha,alpha,0)\mathbf{E}^{\prime}=(-\alpha, \alpha, 0)E′=(−α,α,0) and B ′ = ( α / c , 2 α / c , α / c ) B ′ = ( α / c , 2 α / c , α / c ) B^(')=(alpha//c,2alpha//c,alpha//c)\mathbf{B}^{\prime}=(\alpha / c, 2 \alpha / c, \alpha / c)B′=(α/c,2α/c,α/c), which are the electric and magnetic fields as measured by observer B B BBB.
The electric and magnetic fields E ′ = ( − α , α , 0 ) E ′ = ( − α , α , 0 ) E^(')=(-alpha,alpha,0)\mathbf{E}^{\prime}=(-\alpha, \alpha, 0)E′=(−α,α,0) and B ′ = ( α / c , 2 α / c , α / c ) B ′ = ( α / c , 2 α / c , α / c ) B^(')=(alpha//c,2alpha//c,alpha//c)\mathbf{B}^{\prime}=(\alpha / c, 2 \alpha / c, \alpha / c)B′=(α/c,2α/c,α/c) imply that the electromagnetic field strength tensor as seen by observer B B BBB is given by
(3.648) F ′ = ( 0 α − α 0 − α 0 − α 2 α α α 0 − α 0 − 2 α α 0 ) (3.648) F ′ = 0 α − α 0 − α 0 − α 2 α α α 0 − α 0 − 2 α α 0 {:(3.648)F^(')=([0,alpha,-alpha,0],[-alpha,0,-alpha,2alpha],[alpha,alpha,0,-alpha],[0,-2alpha,alpha,0]):}F^{\prime}=\left(\begin{array}{cccc} 0 & \alpha & -\alpha & 0 \tag{3.648}\\ -\alpha & 0 & -\alpha & 2 \alpha \\ \alpha & \alpha & 0 & -\alpha \\ 0 & -2 \alpha & \alpha & 0 \end{array}\right)(3.648)F′=(0α−α0−α0−α2ααα0−α0−2αα0)
The electromagnetic field strength tensor as seen by observer C C CCC is then given by F ′ ′ = Λ F ′ Λ T F ′ ′ = Λ F ′ Λ T F^('')=LambdaF^(')Lambda^(T)F^{\prime \prime}=\Lambda F^{\prime} \Lambda^{T}F′′=ΛF′ΛT, where
(3.649) Λ = ( γ − β γ 0 0 − β γ γ 0 0 0 0 1 0 0 0 0 1 ) (3.649) Λ = γ − β γ 0 0 − β γ γ 0 0 0 0 1 0 0 0 0 1 {:(3.649)Lambda=([gamma,-beta gamma,0,0],[-beta gamma,gamma,0,0],[0,0,1,0],[0,0,0,1]):}\Lambda=\left(\begin{array}{cccc} \gamma & -\beta \gamma & 0 & 0 \tag{3.649}\\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)(3.649)Λ=(γ−βγ00−βγγ0000100001)
Here β = β ( v ) ≡ v c β = β ( v ) ≡ v c beta=beta(v)-=(v)/(c)\beta=\beta(v) \equiv \frac{v}{c}β=β(v)≡vc and γ = γ ( v ) ≡ 1 1 − β 2 = 1 1 − v 2 c 2 γ = γ ( v ) ≡ 1 1 − β 2 = 1 1 − v 2 c 2 gamma=gamma(v)-=(1)/(sqrt(1-beta^(2)))=(1)/(sqrt(1-(v^(2))/(c^(2))))\gamma=\gamma(v) \equiv \frac{1}{\sqrt{1-\beta^{2}}}=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}γ=γ(v)≡11−β2=11−v2c2. Thus, one obtains
F ′ ′ = Λ F ′ Λ T = ( 0 α − α ( 1 − β ) γ − 2 α β γ − α 0 − α ( 1 − β ) γ 2 α γ α ( 1 − β ) γ α ( 1 − β ) γ 0 − α 2 α β γ − 2 α γ α 0 ) (3.650) = ( 0 − E ′ ′ 1 − E ′ ′ 2 − E ′ ′ 3 E ′ ′ 1 0 − c B ′ ′ 3 c B ′ ′ 2 E ′ ′ 2 c B ′ ′ 3 0 − c B ′ ′ 1 E ′ ′ 3 − c B ′ ′ 2 c B ′ ′ 1 0 ) F ′ ′ = Λ F ′ Λ T = 0 α − α ( 1 − β ) γ − 2 α β γ − α 0 − α ( 1 − β ) γ 2 α γ α ( 1 − β ) γ α ( 1 − β ) γ 0 − α 2 α β γ − 2 α γ α 0 (3.650) = 0 − E ′ ′ 1 − E ′ ′ 2 − E ′ ′ 3 E ′ ′ 1 0 − c B ′ ′ 3 c B ′ ′ 2 E ′ ′ 2 c B ′ ′ 3 0 − c B ′ ′ 1 E ′ ′ 3 − c B ′ ′ 2 c B ′ ′ 1 0 {:[F^('')=LambdaF^(')Lambda^(T)=([0,alpha,-alpha(1-beta)gamma,-2alpha beta gamma],[-alpha,0,-alpha(1-beta)gamma,2alpha gamma],[alpha(1-beta)gamma,alpha(1-beta)gamma,0,-alpha],[2alpha beta gamma,-2alpha gamma,alpha,0])],[(3.650)=([0,-E^(''1),-E^(''2),-E^(''3)],[E^(''1),0,-cB^(''3),cB^(''2)],[E^(''2),cB^(''3),0,-cB^(''1)],[E^(''3),-cB^(''2),cB^(''1),0])]:}\begin{align*} F^{\prime \prime} & =\Lambda F^{\prime} \Lambda^{T}=\left(\begin{array}{cccc} 0 & \alpha & -\alpha(1-\beta) \gamma & -2 \alpha \beta \gamma \\ -\alpha & 0 & -\alpha(1-\beta) \gamma & 2 \alpha \gamma \\ \alpha(1-\beta) \gamma & \alpha(1-\beta) \gamma & 0 & -\alpha \\ 2 \alpha \beta \gamma & -2 \alpha \gamma & \alpha & 0 \end{array}\right) \\ & =\left(\begin{array}{cccc} 0 & -E^{\prime \prime 1} & -E^{\prime \prime 2} & -E^{\prime \prime 3} \\ E^{\prime \prime 1} & 0 & -c B^{\prime \prime 3} & c B^{\prime \prime 2} \\ E^{\prime \prime 2} & c B^{\prime \prime 3} & 0 & -c B^{\prime \prime 1} \\ E^{\prime \prime 3} & -c B^{\prime \prime 2} & c B^{\prime \prime 1} & 0 \end{array}\right) \tag{3.650} \end{align*}F′′=ΛF′ΛT=(0α−α(1−β)γ−2αβγ−α0−α(1−β)γ2αγα(1−β)γα(1−β)γ0−α2αβγ−2αγα0)(3.650)=(0−E′′1−E′′2−E′′3E′′10−cB′′3cB′′2E′′2cB′′30−cB′′1E′′3−cB′′2cB′′10)
Therefore, one finds that E ′ ′ 1 = − α , E ′ ′ 2 = α ( 1 − β ) γ = α c − v c + v , E ′ ′ 3 = 2 α β γ = E ′ ′ 1 = − α , E ′ ′ 2 = α ( 1 − β ) γ = α c − v c + v , E ′ ′ 3 = 2 α β γ = E^(''1)=-alpha,E^(''2)=alpha(1-beta)gamma=alphasqrt((c-v)/(c+v)),E^(''3)=2alpha beta gamma=E^{\prime \prime 1}=-\alpha, E^{\prime \prime 2}=\alpha(1-\beta) \gamma=\alpha \sqrt{\frac{c-v}{c+v}}, E^{\prime \prime 3}=2 \alpha \beta \gamma=E′′1=−α,E′′2=α(1−β)γ=αc−vc+v,E′′3=2αβγ= 2 α v c 2 − v 2 , B ′ ′ 1 = α / c , B ′ ′ 2 = 2 α γ / c = 2 α 1 c 2 − v 2 2 α v c 2 − v 2 , B ′ ′ 1 = α / c , B ′ ′ 2 = 2 α γ / c = 2 α 1 c 2 − v 2 2alpha(v)/(sqrt(c^(2)-v^(2))),B^(''1)=alpha//c,B^(''2)=2alpha gamma//c=2alpha(1)/(sqrt(c^(2)-v^(2)))2 \alpha \frac{v}{\sqrt{c^{2}-v^{2}}}, B^{\prime \prime 1}=\alpha / c, B^{\prime \prime 2}=2 \alpha \gamma / c=2 \alpha \frac{1}{\sqrt{c^{2}-v^{2}}}2αvc2−v2,B′′1=α/c,B′′2=2αγ/c=2α1c2−v2, and B ′ ′ 3 = α ( 1 − β ) γ / c = B ′ ′ 3 = α ( 1 − β ) γ / c = B^(''3)=alpha(1-beta)gamma//c=B^{\prime \prime 3}=\alpha(1-\beta) \gamma / c=B′′3=α(1−β)γ/c= α c c − v c + v α c c − v c + v (alpha )/(c)sqrt((c-v)/(c+v))\frac{\alpha}{c} \sqrt{\frac{c-v}{c+v}}αcc−vc+v. Summarizing, the electric and magnetic fields as measured by observer C C CCC are
(3.651) E ′ ′ = ( E ′ ′ 1 , E ′ ′ 2 , E ′ ′ 3 ) = ( − α , α c − v c + v , 2 α v c 2 − v 2 ) (3.652) B ′ ′ = ( B ′ ′ 1 , B ′ ′ 2 , B ′ ′ 3 ) = ( α c , 2 α 1 c 2 − v 2 , α c c − v c + v ) (3.651) E ′ ′ = E ′ ′ 1 , E ′ ′ 2 , E ′ ′ 3 = − α , α c − v c + v , 2 α v c 2 − v 2 (3.652) B ′ ′ = B ′ ′ 1 , B ′ ′ 2 , B ′ ′ 3 = α c , 2 α 1 c 2 − v 2 , α c c − v c + v {:[(3.651)E^('')=(E^(''1),E^(''2),E^(''3))=(-alpha,alphasqrt((c-v)/(c+v)),2alpha(v)/(sqrt(c^(2)-v^(2))))],[(3.652)B^('')=(B^(''1),B^(''2),B^(''3))=((alpha )/(c),2alpha(1)/(sqrt(c^(2)-v^(2))),(alpha )/(c)sqrt((c-v)/(c+v)))]:}\begin{align*} & \mathbf{E}^{\prime \prime}=\left(E^{\prime \prime 1}, E^{\prime \prime 2}, E^{\prime \prime 3}\right)=\left(-\alpha, \alpha \sqrt{\frac{c-v}{c+v}}, 2 \alpha \frac{v}{\sqrt{c^{2}-v^{2}}}\right) \tag{3.651}\\ & \mathbf{B}^{\prime \prime}=\left(B^{\prime \prime 1}, B^{\prime \prime 2}, B^{\prime \prime 3}\right)=\left(\frac{\alpha}{c}, 2 \alpha \frac{1}{\sqrt{c^{2}-v^{2}}}, \frac{\alpha}{c} \sqrt{\frac{c-v}{c+v}}\right) \tag{3.652} \end{align*}(3.651)E′′=(E′′1,E′′2,E′′3)=(−α,αc−vc+v,2αvc2−v2)(3.652)B′′=(B′′1,B′′2,B′′3)=(αc,2α1c2−v2,αcc−vc+v)